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Some Basic Concepts of Chemistry in One Shot - JEE/NEET/Class 11th Boards || Victory Batch — Transcript

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0:00hello my shimmering stars i welcome you
0:03all to this awesome platform of pw
0:06english
0:07students as you all know this is victory
0:11batch so i am shorya your chemistry
0:14teacher and we are going to start our
0:17chapter some basic concepts of chemistry
0:22students as victory batch says
0:26the harder the better
0:29the sweeter the victory so now with this
0:33quote we will start our first topic
0:37students as the names suggest some basic
0:41concepts of chemistry
0:44so i would say that there is a word
0:47basic in this
0:49this defines us and reminds of of those
0:52classes the previous classes which we
0:55have started
0:56like class 10th in class 10th we have
0:59studied about matter in our surroundings
1:03so from that portion we are going
1:06towards our 11th portion from 10th
1:09towards 11th so i am going to tell you
1:12every detailed basic detail
1:15of the chapter
1:17first of all the thing that arises in
1:20everyone's mind is the subject chemistry
1:24students chemistry is the branch of
1:28science that
1:30deals with the study of composition
1:33structure
1:35and properties of matter
1:38and also about the changes that matter
1:42undergoes in different conditions
1:45and the various laws that govern those
1:48changes
1:50so students chemistry plays a vital role
1:54in science
1:55students it plays a vital role in health
1:58care products
2:00applications in war applications in
2:04industry
2:05applications in food
2:07so in our day-to-day life chemistry has
2:11an important role
2:13so now moving towards this in our
2:15surroundings when we look around we look
2:18around a lot of things
2:20a lot of objects the most common object
2:23which a student observe in its classroom
2:27is the duster
2:29well
2:30duster is the kind of solid object right
2:34okay now let me what uh let me tell what
2:36is the glass of water
2:38glass of water is a liquid right so
2:41water comes under the category of liquid
2:43phase now students these substances
2:46which we look around in our surroundings
2:49are all matter
2:51now what is matter
2:54matter is a substance
2:56that possesses mass
2:59and occupies space
3:03one more thing that defines this more
3:06accurately i would say is we can
3:09eventually feel with our senses right so
3:12what would be the definition of matter i
3:15would say
3:16matter it is a substance
3:20it is
3:21a substance
3:27which possesses
3:30which
3:31possesses mass
3:35and
3:37occupy
3:39space
3:41and occupy space
3:43and
3:44can be felt
3:46can be
3:48felt
3:49by our
3:52senses
3:54so as i told you example of duster of
3:58water we can take more examples like
4:00chalk we have gases around us all these
4:05are matter students
4:07you know what matter can be classified
4:10into two basic categories the
4:13number first one is the physical
4:15classification and the next one is the
4:18chemical classification so when we
4:20classify them
4:22when we
4:23classify matter
4:26classification
4:28it is
4:29the
4:30physical one
4:32and
4:33the chemical one students when we study
4:37about physical classification you know
4:39physical classification contains it
4:42tells us about states of matter now what
4:45are states of matter the most common and
4:48the famous example is in front of you
4:51this is solid liquid and gas
4:54students uh when we take a ice cube from
4:58our freezer in our home in our kitchen
5:00we take it out and we put it into a bowl
5:03it starts melting
5:05when that liquid is kept on a stove and
5:08is heated it converts into vapors so
5:12what is this these are states of matter
5:15the ice cube that we took from the
5:16freezer is a solid
5:19like the water in which it converted was
5:23liquid and when then it was heated it
5:26converted into water vapors that water
5:28vapors was gaseous form so what do we
5:31observe students we observe these states
5:33of matter in our day-to-day life
5:36right solid liquid and gas now moving
5:39towards the physical classification in
5:41detail first of all we'll learn about
5:44solid
5:46first of all we will learn about solid
5:49if i talk about solid solid is an object
5:54in which
5:55what we can observe
5:57we can observe that particles in solid
6:00are so tightly packed for example
6:04this is my first particle this is my
6:06second third and fourth this is the
6:08first layer now how is the second layer
6:12formed here the second layer is
6:14similarly just like the first layer they
6:16are so tightly packed with each other
6:20like this way one two
6:22three
6:23four now the third layer is in such a
6:27form
6:28now what you can see by
6:30just being
6:32looking at the example i can observe
6:34that there are some particles let us say
6:36these are a particles or b particles
6:39okay let us say a b
6:41a b kind of particles if i say a b a b
6:45kind of particles they are so tightly
6:48packed this is my a particle this is my
6:50b particle they are so tightly packed
6:53that they cannot move from their mean
6:55position they will only vibrate along
6:58its mean position like two and through
7:00just like this they can't move from one
7:02position to the another position and
7:04what is the reason behind this that they
7:07have a strong forces of attraction
7:10do remember that so the first point
7:13which we need to write here is the
7:15particles are
7:17the particles are
7:20tightly packed
7:22particles are tightly bagged students
7:25the next point
7:27which we have to remember that there is
7:30a strong forces of attraction moreover
7:33we can say they have a definite shape
7:34and definite volume that these are the
7:37basic things definite shape definite
7:39volume we have started these in class
7:4110th right but 11th the most important
7:44cheese important thing you need to
7:46remember is the force of attraction and
7:49the force of repulsion right so next
7:51point is force of attraction
7:55force of attraction is very very strong
8:00due to which they are unable to move
8:02students
8:04due to which they are unable to move now
8:07coming towards the third most point i
8:09would say
8:10in the case of solids there is very very
8:14less
8:15forces of repulsion or i would say the
8:17tendency to escape each other
8:20how to escape each other if there if one
8:22particle is escaping from another
8:24particle they attain a distance they are
8:28at a far most distance and if there is a
8:30strong force of attraction they are
8:32attracted and they are close to each
8:33other in solid they
8:36have zero very very
8:39less forces of repulsion
8:42so
8:44forces of repulsion which i already told
8:47you to maintain distance between each
8:49other
8:50right
8:51in case of solid it is very very less
8:54which is almost negligible students okay
8:58this was all about solid you can even
9:00add more points like definite shape and
9:01definite volume okay now moving towards
9:04the next case the next case is our
9:07liquid
9:09okay
9:10if i talk about liquid now let us see
9:13its structure first of all in the case
9:15of liquid the particles are at a
9:17distance from each other but not at a
9:20far most distance they are
9:22at a little bit distance from each other
9:24for example if i see this is my first
9:26particle then this is the second one
9:28third one fourth one fifth one sixth one
9:30seventh one eighth one so what are the
9:32things that i'm observing here students
9:35i am observing
9:37that number first
9:39the particles are
9:44quite at a distance from each other
9:50but not far
9:54they are close but not far but at a
9:56distance from each other the second
9:58point which we observe here is the
10:00forces of attraction see in the case of
10:02liquid both the forces act forces of
10:04attraction and forces of repulsion there
10:07is a forces of attraction that's why
10:08they are attracted and there is forces
10:11of repulsion that's why they are not so
10:13closed attached they are at a little
10:15distance because both the forces act so
10:17i would say the forces of attraction
10:21forces of attraction in the case of
10:23liquids
10:26is less than solids
10:28but
10:31more than gas
10:32more than
10:33gas
10:36right okay
10:38now coming towards the third most point
10:40that is the forces of repulsion so if i
10:42talk against gases i would say
10:44liquid you know has a less force of
10:47repulsion as compared to gaseous form
10:50and i would say it has more uh repulsion
10:53as compared to the solid form so the
10:55third most point is
10:59force of repulsion
11:02is
11:04more than solids
11:10but
11:11less than
11:13liquids
11:15okay so students this was all about
11:18liquid state okay i would say that they
11:21um have a definite volume you know but
11:24they don't have a definite shape for
11:25example if i take a beaker you know and
11:28if i transform that weak uh the you know
11:31liquid from one substance into another
11:32substance it would occupy its uh you
11:35know shape so shape is not certain for
11:38the liquid case
11:39hence its volume is a certain that means
11:42its volume is definite but not the shape
11:44one right so the third thing that we
11:46talk about is our gaseous form
11:50is our gaseous form if i talk about
11:53gaseous form first of all let us draw
11:55its structure
11:56and while drawing its structure
11:58we can easily tell the particles that
12:01are at a foremost distance for example
12:03this is the first particle it's the
12:04second it's the third like this they are
12:06at a distance so you can easily observe
12:09these things like
12:10solid in they are very close in liquid
12:12they are at a distance but in gaseous
12:14they are at the far most distance from
12:16each other so when they are at a far
12:18most distance from each other first
12:19thing we can easily say that they have a
12:22very strong forces of repulsion and how
12:25are the forces of attraction they are
12:27very less very less so first of all the
12:31very first point which we are going to
12:32write particles are far
12:37very far
12:40right the second most point students
12:42that i am going to write here is the
12:44forces of attraction
12:50so the forces of attraction
12:52is very less
12:54very very less
12:56and the very last point students i would
12:58say that the forces of repulsion
13:04so the forces of repulsion are very
13:07strong
13:08are very strong
13:10so students this was all about the
13:13physical classification right so if i
13:16say about the definite shape and
13:18definite volume in the case of gases
13:19there is no definite shape no definite
13:22volume um in the case of gaseous right
13:25so this was all about the physical
13:26classification of matter now we are
13:28proceeding towards the chemical
13:30classification of matter but before that
13:32let me tell you that in the case of
13:34physical classification we observe that
13:36these three forms these three states of
13:39matter are interconvertible that means
13:41that we can say that solids
13:45can be converted into liquids liquids
13:49can be converted into gases
13:51and gases can be converted into solids
13:54and vice versa how
13:57how i am saying that liquids can further
14:00back be converted into solids solids can
14:02further back be converted into gaseous
14:04and gaseous can further back be
14:07converted into liquid how can this be
14:09possible in class 10th we have already
14:11studied that while heating and cooling
14:13we observe such situations you know when
14:16i take ice from the freezer and keep it
14:18in a bowl it converts right into what
14:21into water when i boil that water it
14:23converts into what it converts into
14:25gaseous form okay so these are all
14:28things we need to understand that these
14:30three states of matters are
14:31interconvertible into each other
14:33now students coming towards the chemical
14:36classification of
14:38matter
14:39this is our
14:41chemical classification
14:44okay if i talk about chemical
14:46classification the very first thing
14:48arises is that it is classified into two
14:51basic categories one is pure another one
14:53is mixture so by looking at the words
14:55you can easily identify what do you mean
14:57by pure and what do you mean by mixture
14:59mixture is something in which
15:00something is mixed right and pure is
15:03something which is pure and it and
15:06something is not added into it right so
15:08these are the basic things when we
15:10listen to a word and we could you know
15:12um think about it so first of all i
15:14would say that
15:16when
15:17we classify matter according to the
15:19chemical chemical classification we
15:21observe the pure substances pure
15:23substances are further classified into
15:25elements and compounds and students
15:27elements are
15:29the one of the kind how i'm saying that
15:31in elements you can see only
15:35one kind of matter
15:37only one kind of atom
15:41how i can say that example carbon
15:44example hydrogen okay example oxygen you
15:47can even write that only one kind
15:50only carbon only hydrogen only oxygen
15:53only sodium only magnesium only one kind
15:56of atom when we observe that that is
15:59referred to as an element and element
16:01students further is classified into
16:03three basic categories first is metals
16:05second is non-metals and third one is
16:08students metalloids so i would say that
16:10if i classify them into elements then it
16:13is classified into metals
16:15the second one is
16:17non-metals
16:19students and the third one is metalloids
16:23now what are metals non-metals and
16:25metalloids if i speak about metals to
16:27you metals are the ones that are good
16:30conductor of heat and electricity metals
16:32are the ones that are lustrous you have
16:34studied in class 10 to the chapter
16:36metals and non-metals well
16:38that is what metal is it is lustrous it
16:41is good conductor of heat and
16:42electricity you know it is malleable it
16:45is ductile what is malleable and ductile
16:47malleable are the ones that can be
16:48hammered into thin sheets and ductile
16:51are the ones that can be converted into
16:53wires for example copper magnesium iron
16:57these are metals right
16:59now coming towards non-metals i would
17:01say non-metals are the one that are poor
17:03conductor of heat electricity you know
17:05so they are not malleable not tactile we
17:08observe these things in non-metals while
17:10coming across towards metalloids
17:12metalloids are the ones that exhibit the
17:14properties of both metals as well as
17:17non-metals so these were the basic
17:19things that we started in the class you
17:21know tenth if i take example of
17:23metalloids i would say that metalloids
17:24can be
17:26anti-bismuth arsenic right so this was
17:29all about elements if i come towards the
17:31second category of pure substances that
17:33is compounds compounds are the one which
17:36contain two or more than two atoms you
17:39know compounds
17:42compounds they are the one
17:44which contain two
17:47or
17:48more than
17:51two
17:52kinds of atoms see
17:55if you find only a single atom that is
17:59known as element but when you find two
18:02or more than two elements which are
18:04combined together they are known as
18:06compounds students compounds can be
18:08further categorized into organic and
18:10inorganic
18:16they can be further classified into
18:18organic and inorganic
18:20organic compounds are those in which we
18:23study about hydrocarbons now what is
18:25hydrocarbons you will observe the
18:27hydrocarbons chapter in class 11th only
18:30hydrocarbons are the one in which we
18:33observe carbon as well as hydrogen this
18:35is the basic necessity condition carbon
18:37and hydrogen should be present other
18:39than that any other element can also be
18:41there you know
18:43but if only carbon is present it cannot
18:45be categorized into organic compounds do
18:47remember that so organic compounds are
18:50basically derived from
18:53animals and plants but if i talk about
18:55inorganic compounds inorganic compounds
18:57are the one that derived from minerals
18:59and rocks clear so this was all about
19:02the compounds category now moving
19:03towards the mixture as i know mixture
19:06itself tells us that something is mixed
19:09you know it is further classified into
19:11homogeneous and heterogeneous as we are
19:14assigned student students whenever you
19:17are studying any of the topic do
19:19remember that
19:21kindly take the word if it is splitting
19:23into two parts do that it will make your
19:26concept easy and you will remember it
19:29here and you'll never forget it you know
19:32some people what did what they do
19:34they just you know
19:36keep keep again and again again and
19:38again writing and learning and writing
19:39and learning and after few times they
19:41just forget it when you understand any
19:44concept you grasp it properly you know
19:47and then and then you become stronger
19:50and then you remember that so it will
19:52help you a lot so try to understand
19:54things instead of you know uh cramming
19:56it so first thing is homogeneous mixture
19:59as i told you please split it in two
20:01parts if i split it into two parts i
20:04observe homo homo refers to what same
20:08homo refers to same and if i speak about
20:11heterogeneous hetero refers to different
20:15now what i
20:16meant by same and different
20:18see if something is present in uniform
20:21you know uniform concentration or same
20:23concentration throughout that is known
20:25as homogeneous mixtures and if something
20:28is present in different composition
20:30throughout that is referred to as
20:31heterogeneous right if i talk about
20:34homogeneous mixture one of the examples
20:35which we observe is basically sugar in
20:38water salt in water okay
20:41example is
20:44sugar
20:45in water
20:47the next one is salt
20:50in water or air i would say
20:53why air i am saying because air is
20:55basically comprised of gases which are
20:57present in uniform concentration
20:59throughout that that's why it lies under
21:01the category of homogeneous mixtures if
21:03i talk about heterogeneous mixture one
21:05of the most common example is smoke
21:07other than that is oil in water right so
21:10heterogeneous we have two cases like oil
21:12and water
21:14where the concentrations are not uniform
21:17the second most example i told you is
21:19smoke so students this was all about our
21:23chemical classification of matter okay
21:26now moving towards the next topic that
21:28is the properties of matter
21:30now we have studied about the
21:32classification of matter now comes the
21:34properties
21:36now comes the
21:38properties of matter students if i talk
21:41about properties of matter what is the
21:43basic thing that i observe here
21:47i see that
21:49the properties of matter can be of two
21:51types one is the physical property
21:53another one is the chemical property
21:55okay
21:57one is
21:59the physical property
22:06okay the second one students
22:09is
22:10the chemical property
22:13okay if i talk about students physical
22:16property what are physical properties
22:18physical properties are the ones which
22:20we can observe and in which the identity
22:23is not lost in which there is no
22:26chemical reaction
22:27okay you can see that the substance
22:30remains the same throughout so i would
22:32say that please do write it or you can
22:34underline in your ncrts physical
22:37properties these are the ones
22:43which do not
22:45which
22:47do not
22:48lose their identity
22:51lose
22:52their identity
22:57and
22:58no chemical reaction
23:01no
23:03chemical
23:05reaction
23:08occurs
23:10here
23:12occurs here
23:14okay example i would say density
23:18volume
23:21boiling point
23:24melting point these are the examples of
23:26the physical properties now coming
23:28towards the next one that is the
23:29chemical properties chemical properties
23:31are the one in which reactant is being
23:33converted into the product that means
23:35the chemical reaction is taking place
23:37and when a chemical reaction is taking
23:39place students identity is lost of the
23:42reactant so i would say the chemical
23:44properties
23:45are the ones
23:49in which
23:52chemical reaction occurs
23:59and they lose
24:02and they lose
24:05their identity
24:11okay
24:12when they are being consumed so students
24:15these are all about a chemical
24:17properties in which you can add a
24:19chemical reaction occurs right okay so
24:22students this will all this was all
24:25about the properties of matter now how
24:28can we miss measurement how can we do
24:31the measurement of these properties of
24:34these physical properties basically
24:35which we can observe which we can think
24:38without any chemical change right so if
24:40i see here about the physical properties
24:43we talk about the measurement and how is
24:45measurement done for example if i say
24:47that i am living in my home and i wanted
24:50to eat some fruits or some vegetables so
24:52what i do is i move from a distance i
24:55move 15 you know meters at a distance
24:58and i go to another place and
25:00what i see is there is 15 meters or 15
25:03kilometers i am using 15 here refers to
25:06as a number and kilometers or meters
25:08referred to as a unit so
25:10to measurement of a physical property to
25:13measurementize a physical property we
25:16need to understand the basic term that
25:17is unit a unit is a standard reference
25:21with the help of which we can measure
25:22any of the physical property very easily
25:25like we kept for distance we kept meters
25:28kilometers so that we can calculate them
25:31and we can tell for time we have kept
25:33seconds minutes hours these all are
25:36units which we have kept as a standard
25:39to understand to measure all the
25:41physical properties around us right so
25:43students uh now if i talk about unit
25:46as i have explained you unit can be
25:48further categorized into three basic
25:50categories the number first is the basic
25:53units second one is the fundamental
25:55units and third one is the derived units
25:58if i talk about the basic units there
26:00are basically seven basic units you need
26:03to please uh star mark the situation i
26:05would say
26:07or
26:10you can put a box here
26:12in your ncrts also there are seven basic
26:15units what are these seven basic units
26:17the number first is length the second
26:19one is mass the third one is time the
26:22fourth one is electric current the five
26:25one is thermodynamic temperature next
26:28one is amount of substance and the next
26:31one is luminous intensity so these are
26:33the basic units that we have um to learn
26:37right these seven basic units uh have
26:40particular symbols for them you know how
26:43can we represent them to represent
26:45length we use small letter l to
26:48represent mass we use small letter m to
26:51represent time we use small letter t to
26:55represent electric current students we
26:57use
26:58i it is capital i
27:01to represent thermodynamic temperature
27:04we use capital letter t right to
27:07represent amount of substance this is
27:10very important
27:11it is represented by small n and you are
27:13going to study it it is about mole
27:15concept in this only chapter it is the
27:18most important and most asked topic
27:20small concept so we'll study this there
27:23okay we represent it by small n right
27:26then comes the luminous intensity that
27:28is represented by i
27:30to the base v clear now these all are
27:34given some units like length is given
27:36some unit which is meter which is
27:39referred to as the standard reference
27:41with the help of which we can measure
27:44the quantity right so for length what is
27:46the si unit meter for mass what is the
27:49si unit kilogram for time what is the si
27:52unit second
27:53for electric current what is the si unit
27:56ampere for thermodynamic temperature
27:59what is the si unit kelvin for amount of
28:02stars what is the si unit mole
28:06for luminous intensity what is the si
28:08unit candela students now these units
28:12also have a particular symbols we don't
28:14write the big big names we just use the
28:16symbols to represent them for meter the
28:19symbol is small m for kilogram the
28:21symbol is kg for second the symbol is
28:25only small letter s for ampere the
28:27symbol is capital a for kelvin the
28:31symbol is capital k for mole it is m o l
28:35and for candela it is cd so these were
28:38the seven basic units that you need to
28:40understand students now comes the second
28:42category that is the fundamental units
28:46that is
28:50the
28:51fundamental units now what are
28:54fundamental units students fundamental
28:57units are the ones that cannot be
28:59derived from any another units they do
29:02not depend on any of the unit you know
29:06they are
29:07independent
29:14can't be derived
29:20can't be derived we cannot derive these
29:23from any another units from any units
29:30so these are the fundamental units
29:32fundamental units are basically the unit
29:35of mass the unit of time the unit of
29:37length right so now next comes the types
29:40of unit which are referred to as the
29:43derived units derived units are the one
29:46that can be derived you know from the
29:48basic units for example if i take an
29:50example of area students if i talk about
29:54area if i consider about square what i
29:56see square has sides equal sides right
30:00and if i say that
30:01each side has a length l
30:04right or you can say a when we consider
30:07this in mathematics don't worry in
30:09mathematics there are uh some concepts
30:11that students uh feel that it's very
30:13tough and in chemistry also numerical
30:15portion ma'am how we'll do that our
30:17maths is very weak so don't worry i am
30:20there for you and the maths which you
30:22need in chemistry i am going to explain
30:25you
30:25so no need to worry just you need to
30:28understand it need to make notes by hand
30:31first listen from me and then write
30:34accordingly and just practice practice
30:37practice because when you practice that
30:39practice will leads you towards the
30:42victory you know clear so practice is
30:45the most important
30:47right now see i'm talking about area and
30:50if i talk about square what is the
30:52formula of area it is length into length
30:55and now only we have started you can see
30:57in this case
30:58length length is one of the basic unit
31:00and when i multiply one basic unit into
31:03another basic unit that means i am
31:05deriving something so hence it comes
31:08under derived units are you
31:10understanding if i am only talking about
31:12length that is our basic unit but if i
31:15am talking about length into length that
31:18means i'm deriving something and where
31:19i'm deriving something from the basic
31:21units they are referred to as derived
31:23units so here one of the most another
31:25example i would say is volume
31:27in the case of volume what is the
31:29formula length into length into length
31:32right yeah otherwise i would say length
31:34into breadth into height all you can say
31:36is this right so that is volume here
31:38also it is derived why it is derived
31:40because length is a basic unit and when
31:41we multiply one basic unit into another
31:43into the third one it is the derived
31:45units now let us read this what is given
31:48here the units of all other physical
31:50quantities are derived from those of
31:53above quantities which we have started
31:55already the right these units thus
31:58obtained these are those obtained our
31:59derived units area volume these are now
32:02derived units some common physical
32:04quantities and the derived units are
32:06also given to you in this following
32:08table now let us see that the number
32:10first is area which i've already told
32:12you
32:13students it has a formula what is the
32:15formula length into length now coming
32:18towards volume i already told you its
32:20formula is length into length into
32:23length now coming towards density the
32:26most easy formula which we have started
32:29in the small classes and if you have not
32:31started no need to worry you are
32:33studying now right so density formula is
32:35mass by volume right what's its formula
32:39mass
32:40divided by
32:42volume
32:43right okay students now let us see the
32:46next case that is velocity
32:49velocity is distance upon time you have
32:51studied
32:52it in physics also in class 10 so it is
32:55distance
32:59upon time right okay coming towards the
33:02next acceleration to derive acceleration
33:04first you need to calculate velocity so
33:06when you calculate velocity just divided
33:08by time you will get acceleration so
33:11what is the formula of this now
33:14okay its formula is
33:17velocity
33:19divided by
33:20time okay now coming towards the next
33:23part that is our force what is force
33:26students force is refers to um force is
33:28basically refers to as mass into
33:31acceleration right
33:33forces our
33:36mass into acceleration and if i talk
33:38about pressure pressure is force upon
33:41area
33:42so first you need to calculate force for
33:44that right
33:48okay now if i talk about the units here
33:50first of all if i talk about area as you
33:53know length into length first tell me in
33:56the basic unit now we have seen that
33:58length has a si unit which is basically
34:01meter length has an si unit which is
34:04meter right so si unit the international
34:07system of units we have started just
34:08right now the seven basic units are
34:10those right so meter into meter all i
34:13would say is meter square right the very
34:16first one is meter
34:20square if i talk about volume now
34:23see meter into meter into meter what is
34:25this now meter into meter into meter is
34:27cubic meter right so it has
34:31unit as cubic meter
34:34now students coming towards the density
34:36mass by volume first you need to
34:37understand mass mass we have already
34:40started in the basic unit it is kg its
34:42unit is a unit is kg and volume its unit
34:45we have already calculated cubic meter
34:47so hence we can write here
34:52kg
34:53per
34:55cubic
34:56meter so now coming towards velocity it
34:59is distance upon time that means we need
35:02to write the distance s i unit the
35:04distance s i unit is again we know that
35:07it is the meter like if i say i have
35:09traveled this much meter so that is our
35:12meters and then uh that if i talk about
35:16time it is second right so
35:18if i say you hear
35:23okay meter
35:26per second now as you all know
35:28acceleration is velocity by time and we
35:30have already calculated the unit for
35:32velocity it is meter per second and time
35:34is already second so first we'll write
35:36velocity now it is per time so again
35:39time
35:39that means again seconds right now
35:42coming towards force forces mass into
35:44acceleration and we know that mass is
35:46what kilogram and what is acceleration
35:48acceleration is meter meter per second
35:51per second so in short we can write it
35:53is newton it is newton
35:59and if i talk about pressure it is
36:01pascal
36:03right so these were the units now for
36:05these units we have two students write
36:08some symbols okay for area we write a
36:12symbol like for meter square its unit is
36:14meter square to represent it we don't
36:17write such big names we just use symbols
36:19to representation right for meter square
36:21we will write meter square for volume
36:24its cubic meter then meter cube next is
36:27kg per cubic meter then we will write kg
36:31per cubic meter that means meter power
36:33minus 3 right after the students that is
36:37our meter per second the next one is
36:40velocity which is meter per second so we
36:42are going to write meter
36:44second power minus 1 then comes our
36:47acceleration acceleration is meter per
36:49second per second so we have to write it
36:52meter
36:54per second minus 2 right then next one
36:57which arises is force students force is
37:00newtons so we will write it with capital
37:02n and pascal we will write it with pa so
37:05these were the symbols of these derived
37:08units which we have calculated right so
37:11these were all about derived units now
37:13come the prefix in si unit what happens
37:16students when we you know do some
37:18questions we observe that there are some
37:21numbers which are seen in decimal in
37:23fraction for these numbers we use some
37:25certain prefixes for the s i units for
37:28example in some cases you have seen that
37:31there is somewhere written
37:33millimeter nanometered decameter so what
37:37are these these are the prefixes that
37:38are used for decimal fractions all these
37:41cases right so the very first which you
37:43observe which you have seen for 10
37:45raised to the power minus 24 we write it
37:47as y octa and the symbol is used as y
37:50for 10 raised to the power minus 21 we
37:52use prefix zepto and the symbol is z see
37:55you have to just read it once that's it
37:58it is not like you have to cram it okay
38:00no no no not at all you just have to
38:02read this once twice thrice that's it
38:04and the ones that are most important i
38:07will tell you to put a star mark on it
38:09right the next one is 10 to the power
38:11minus 18 which is ato symbol is small a
38:14next is 10 raised to the power minus 15
38:16which prefixes femto and the symbol is
38:18small f the next one is 10 raised to the
38:20power -12 whose prefix is pico and the
38:24simple is small p the next one is 10
38:26raised to the power minus 9 whose prefix
38:29is nano and the symbol is small n
38:32which is important as i told you i will
38:34tell you to mark this one the next one
38:37is 10 raised to the power minus 6 which
38:39is known as micro and the symbol is
38:41mu okay this one is also important the
38:43next one is 10 raised to the power minus
38:453 which is represented as milli and its
38:48symbol is small m while the next one is
38:5110 raised to the power minus 2 and the
38:53prefix is centi symbol is small c the
38:56next one is students
38:58known as deci deci is 10 raised to the
39:01power -1 symbol is small d so these are
39:04important which i have been marking star
39:06for you right now coming towards the
39:08next of the cases we have only 10 whose
39:11prefix is deca which is represented by d
39:14a the next one is 10 raised to the power
39:172 which is hector symbol is h
39:20then is 10 raised to the power 3 which
39:22is kilo symbol is k then is 10 raised to
39:25the power 6 which is mega which is
39:28capital m
39:30next is 10 raised to the power 9 which
39:32is giga which is capital g
39:34next is 10 raised to the power 12 which
39:37is prefixes tetra represented by capital
39:40t the next one is 10 to the power 15
39:43prefix is beta represented by p next is
39:4610 raised to power 18 x are represented
39:48by capital e the next one is 10 raised
39:50to the power 21 which is zeta
39:52represented by capital z and the next
39:55one is 10 raised to the power 24 which
39:57is deuter represented by y so students
39:59you don't have to learn it i'm again
40:02saying just read it once twice or thrice
40:04that's it and from ncrt you can directly
40:06mark the star once that's it clear so
40:09these were all about the derived units
40:12hence we have covered the topic units
40:14now students coming towards the
40:16uncertainty in measurement what is
40:18measurement we have started how to you
40:21know the
40:22properties of measurement and the
40:24measurement of physical properties we
40:26have all covered this topic now coming
40:28towards the uncertainty uncertainty
40:30refers to as if i say some number is
40:321.98 1.96 in decimals or in fractions
40:36that is uncertain what is certain like a
40:38particular number one two three four
40:40five six seven so uncertainty something
40:42that is something is wrong in this you
40:44know
40:44there is some uncertain that measurement
40:47so
40:48here two terms which we need to
40:49understand is precision and accuracy
40:52which we are
40:53going to explain you in a very funny
40:55manner you know um now i will take you
40:58towards your childhood you know okay
41:01how many of you students have played
41:04dark you know dark board game what was
41:07that
41:08let me tell you let us draw it first
41:13okay
41:14let us draw it first
41:16it looked like this you know
41:18it looked like this one
41:22let us make it and let us understand it
41:24in a very beautiful manner students
41:27right
41:30i guess you all have played it because i
41:32have played a lot
41:35i will tell you the rules for this game
41:37now we will learn chemistry with ease in
41:40a very beautiful manner right okay
41:55now let us make it
41:57right
42:05okay your goal is to
42:07put your stick
42:09or the object which you are throwing
42:11into the yellow into the sorry white
42:13portion you have to do what you have to
42:16take a stick which has a magnet in front
42:17of it and this board also has a magnet
42:20in it so what happens is when you throw
42:23it in such a situation it arises just at
42:25the middle and you win the green game
42:28this is the basic rule of the game and
42:30if it arises in you know other regions
42:32of your game board then you are not the
42:35winner okay so what you need to do is
42:37you just need to throw your stick from a
42:39distance and it should arise like this
42:41stuck
42:42okay clear so there what uh how do we
42:45calculate this let me tell you
42:48let us take one more here
42:50okay one the last one
42:53one second students okay
42:58yes
42:59it is almost done so i have made my game
43:03here for you people so now let us learn
43:06it how to play some basic rules here
43:09if you put your stick into the yellow
43:13portion you know you will get five marks
43:16okay if you put your stick into the blue
43:18you will get your four marks if you put
43:21your stick into the pink portion you
43:23will get three marks if you put into the
43:26green one you will get the two marks and
43:28if you put into the first one you will
43:29get the one marks so this is the basic
43:32thing that you need to know in this game
43:34you have to put your stick like this
43:36you know do like this and tuck
43:39okay
43:40make an angle from the eye to put into
43:42the situation now there are two students
43:45that are playing this game okay
43:48so one is me and one is you my students
43:50i would write here
43:52okay one is me
43:55and one is you we both are playing this
43:59game now okay during the class session
44:01so what we are uh taking this stick and
44:04we will put this uh put this stick four
44:07times and we will observe who is the
44:09winner so i'm writing here case one
44:14case two
44:16case
44:17three
44:19case four okay students let us see who
44:22wins the situation right okay now in
44:25this game students you have taken your
44:29stick into your hand and you are going
44:32to
44:32do a blast here let us see where you are
44:35going to put you have put your stick
44:37into number
44:39one
44:41you have put your stick into number one
44:44okay now it's my turn you know so as i'm
44:48doing i'm not so good in this game so
44:49but i'll try my best to defeat you guys
44:52see so i have put my stick and i'm
44:55putting my stick my stick is of red
44:56color yours is one is of white color
44:58right i am putting my stick into
45:02the fourth one okay my stick is into the
45:05fourth one remember one thing when your
45:08stick arises the first one you are the
45:10winner but there are four chances that
45:12we are getting and we will observe these
45:15chances right now the case too again you
45:17have taken your stick and what you are
45:19doing you are again going to put it
45:23right when you are again going to put it
45:25what you have done it arises into the
45:27second most number okay
45:29second most number now again it is my
45:32turn i am doing this again again again i
45:35put into four
45:37see i don't know how to play it because
45:38i have not practiced it it's long time i
45:41played in childhood right so again i
45:43have got fourth number because i put my
45:46slot into the fourth ring now it's your
45:49turn students again and you are full
45:51determined to you know win the situation
45:54you are in the victory batch and you
45:55have to achieve it you know that so what
45:57you are going to do you again put it in
45:59the first box because this is your aim
46:01students
46:02if you are for neat aspiring jesus
46:04parent bored as parent you have to put
46:06your stick here you know right so you
46:09have put again the first one now it's my
46:12turn
46:13so what i'm going to do is i'm going to
46:15put again here
46:17because far at a distance i have not
46:18practiced i'm like throwing like this
46:20you know so again four
46:22no problem my students wins then i will
46:25win automatically i am the winner if you
46:27are the winner right okay the last case
46:30now again you have put your stick into
46:32the first situation well well done
46:35students and again me
46:38i have done here only in the fourth
46:41no problem next time i will try to
46:43defeat you right so this was our game
46:45now who is the winner most probably
46:47which we can see very clearly the person
46:49who hits the center most position is the
46:51winner and you guys are the winner you
46:53have hit your target very easily so one
46:56one one one you have seen but in the
46:58second case you got two so what we have
47:01seen
47:02this situation this situation is known
47:06as accuracy
47:08accuracy coming towards chemistry
47:10students come back come back from the
47:13game your score is one two one one that
47:16means you have to hit one and you have
47:18hit one thrice but once it was twice you
47:21are near to the answer
47:25near to answer close to the answer and
47:27when someone is close to the answer it
47:30is known as
47:31accuracy
47:32now coming towards me who secured four
47:35four four four during this hit now i
47:38have been doing four i was not good at
47:41the point to target it but see what i
47:44did it was my consistency that i
47:47maintained it four four four four four
47:49so this situation this situation is
47:53known as precise
47:56this situation is precision
47:59now what is the situation when there is
48:01a consistency for the same quantity like
48:05for same quantity refers to as four four
48:07four four four so there is a consistency
48:09that is referred to as precision and
48:11when there is a closeness towards your
48:14target that refers to as accuracy
48:17hopefully you love this game a lot right
48:19so let us first read this definitions to
48:21make it more clearly understandable
48:23first of all precision i would talk
48:24about refers to the closures of
48:27various measurements for
48:29same quantity now you will say that mem
48:32for closures or for you know close you
48:35have told the word accuracy when it is
48:37near to the answer but you are using the
48:40word here closure here closure refers
48:42for the same quantity four four four
48:44four four right that means consistency
48:47and if i talk about accuracy it is the
48:49agreement of a particular value to the
48:51true value of the result
48:53true value is 1 and it is the agreement
48:56that it is close to the true value so
48:58that is accuracy clear now let us move
49:00towards a wonderful example of ncrt
49:02there are three students students a b
49:04and c and they have been doing a match
49:07in that they got some measurements and
49:10they participated twice so they got
49:12measurement one and measurement two and
49:14the correct answer students
49:20the correct answer was 2 right now let
49:23us check here
49:25which is accurate and which is precise
49:26right so the student first a let us talk
49:29about student a if i talk about student
49:32a students uh the first time measurement
49:35was 1.95 the next time measurement was
49:371.93 now both the situations you know
49:40are close by but they are not close to
49:42the number they are not close to the
49:45target too hence they are consistent
49:47because the number is almost equal you
49:50can say that ma'am you have got four
49:51four four four four but they are not
49:541.939393 no but at least they are close
49:57to the same quantity closure to the same
50:00quantity that means they are precise
50:01they are consistent so this situation is
50:04precise are they accurate when will be
50:06they accurate they will be accurate when
50:08they are close to the target but target
50:10is two hence they are not accurate so
50:12accuracy is absent in this case right if
50:15i talk about the second situation of the
50:16student b there are two measurements
50:19again the first measurement is 1.94 and
50:212.05 now students please observe firstly
50:24about accuracy accuracy is the one when
50:26we are close to the target the target is
50:28two is nine point one point nine four
50:31and two point zero five close to two no
50:33what is close to close to is like uh two
50:36two is something like one point nine
50:37nine two point zero one these are close
50:39to two hence they are not accurate so
50:41now let us talk about uh precision are
50:44they precise no because there is no
50:46consistency this comes under the
50:48category of one something this con comes
50:50under the category of two something so
50:52here they are not precise at all now uh
50:54if we'll talk about student c what we
50:56observe here 2.01 and 1.99 yes they are
51:00close to two hence they are accurate
51:03so tell us about precision are the
51:05precision yes they are precision
51:07i have told you that one point they
51:09comes under quantity of one two so they
51:11are not precise because they are not
51:13some of you know uh consistency is not
51:15there of the same quantity here it is
51:17almost same quantity two point zero one
51:19point nine nine they are almost nearby
51:22very nearby and that means they have
51:25consistency hence they are precise so we
51:28have learned about accuracy and
51:30precision this is very most important
51:32which you need to understand and which
51:34you need to keep in your minds okay so
51:36now moving towards the next that is
51:38scientific notations students scientific
51:40notations are the one that means when we
51:42represent a number in the form of 10
51:44raised to the power something for
51:46example you have been given any number
51:49like
51:510.00978 you are given this number and i
51:54am saying that please represent this
51:55number in forms of 10 raised to the
51:57power something when i represent this
51:59number in terms of 10 raised to power
52:01something it is referred to as
52:03scientific notations how
52:05in 10 raised to power n n can be 1 2 3 4
52:085 6 7 8 like this right so see how to do
52:11this how to convert it into such
52:13situations there are certain rules which
52:15we need to understand the very first
52:18rule
52:19rule number one
52:24rule number one states that if we shift
52:27if
52:29we shift
52:30towards the right
52:33from the decimal
52:35towards right
52:39from decimal
52:42then
52:45then students
52:47power is
52:49negative
52:51power is negative
52:54example please let us see
52:57example is 0.009
53:01you need to understand the two basic
53:02terms which i already you know i want
53:05you to know so that it becomes easy for
53:08you
53:09okay i'm making a box here in this you
53:11need to write non-zero digits
53:17what are non-zero digits students
53:19non-zero digits are the one that is 1 2
53:233
53:234 5 6 7 8 9
53:28okay which are other than zero they are
53:30referred as nonzero digits now you need
53:33to for to accomplish the rule one you
53:35need to understand what is non-zero
53:37digit which i have told you first find
53:39out that whenever we shift towards the
53:42right from the decimal then we need to
53:43under find the first non-zero digit
53:46here
53:48this nine
53:49is a non-zero digit zero is a zero digit
53:52zero is a zero digit zero is a zero
53:53digit but nine lies in the category of
53:56non-zero digit so when you find the
53:57first non-zero digit what you need to do
53:59is you need to move this decimal after
54:02this
54:03you have to do this to make it bring
54:05under the category of scientific
54:06notation that is 10 raised to the power
54:08something so what you are going to do is
54:10you are moving moving in such a
54:11situation like one decimal place two
54:14decimal place three decimal place and
54:15after this you want to get now so what
54:17you will do here you will do here that
54:21after nine you want point decimal you
54:24will put decimal what comes next 7 it
54:26now you will write into 10 raise to the
54:28power what you are moving in which
54:30direction towards the right hand
54:32direction from the decimal then you have
54:34to write the power in negative form and
54:36if the power is a negative form then how
54:38can you count one two three so after
54:40third position decimal is coming power
54:42is in negative form so 10 raised to the
54:44power -3 clear so now comes the next
54:47rule students
54:50that is our rule number
54:53two
54:54okay if i talk about students the rule
54:57number two i would say that if we shift
55:02if we shift towards
55:05left position
55:09from decimal
55:14from
55:17decimal then power is positive
55:23example
55:25okay i'm taking one more example it is
55:287 000 okay now any decimal present no
55:32decimal present that means we have to
55:34move in from the left position we will
55:37move from the towards the left position
55:39when there is decimal then you have to
55:41move into the right position from the
55:43first non-zero digit what is the first
55:45nonzero digit here 7 so we need to put
55:48decimal after the first non-zero digit
55:49hence after 7 we have to put so we have
55:52to put we will move in this direction to
55:54put it after 7 how many 0's have to
55:56cross 1 to 3 that means 7 into
56:0010 raised to the power what three it
56:03should be positive because we are moving
56:05in which direction we are moving
56:06students in our left direction so this
56:09was all about scientific notations and
56:12in ncrt also you will observe some basic
56:15things for example addition subtraction
56:17multiplication and division the very
56:19first thing that you need to remember in
56:21all all those cases is that the basic
56:23rule when you are given some numbers to
56:25add for example 2.1 i am taking here 2.1
56:29plus 4.22
56:31what you will do this you will add both
56:33of this number when you are going to add
56:34these numbers now students you will get
56:36the answer is 6.3 to write you are
56:39getting this answer this answer is
56:41incorrect why this is incorrect because
56:44you need to find out here what you need
56:47to find out here that during the you
56:49know
56:50during this question it is the simplest
56:53but when you get into the terms of
56:54scientific notations
56:56this is the simplest case when you can
56:58directly add but for example
57:00these decimals are not given you are
57:02given something in 10 raised to the
57:03power something in scientific notation
57:05what you will do for example if i say
57:075.7 into 10 raised to the power 6
57:11okay you need to add it into 4.2 into 10
57:13raised to the power 6.
57:16what you will do
57:18what you will do students when you are
57:19going to add both of these situations
57:21you can directly add them but when you
57:24are given a situation to multiply what
57:26you are going to do in multiply case it
57:28is a quite different situation in
57:30addition subtraction it is directly you
57:32are taking 10 raised to the power 6
57:33common and then you will go to add them
57:36this is the basic way of addition and
57:37subtraction which we do in mathematics
57:39but if i
57:41if i put a symbol of multiplication here
57:44what you will do in multiplication
57:46students the powers are being added
57:48while in division the powers are being
57:50subtracted so when the powers are being
57:52added here what is going to happen here
57:565.7
57:58into 4.2 as i told powers are going to
58:01add so 10 raised to the power 6 plus 10
58:04raised to power 6 can be written as 10
58:06raised to the power 6 plus 6 clear
58:08because powers are already being added
58:11you can multiply this and the answer
58:13will be something into 10 raised to the
58:14power 12 but in the case of division for
58:17example i give you the same situation in
58:19the division case how can you do so
58:22please see that also
58:24it's very simple you will get these all
58:26in mathematics also but still i'm
58:28telling you so that you don't get
58:29confused there are very less chances of
58:31the questions in these kinds of stuff
58:34just you need to know about these two
58:36rules since i'm going to tell you so
58:38that it is more clear to you right so
58:40one more case if i take i if i say
58:47it is like i'm taking 5.6 into 10 raised
58:51to power 6 divided by 4.2 into 10 raised
58:55to the power 4 so what we'll do in the
58:58case of divide divide the powers are
59:01being subtracted like this
59:04and then we are going to write 5.6
59:08divided by 4.2 like the students so here
59:11powers are being subtracted here powers
59:13are being added clear it is very simple
59:16students now students let us see the
59:19very next topic which we are going to
59:21study is significant figures significant
59:24figures is a very interesting topic and
59:25you will find it very easy you know
59:28there are some students who feel
59:29mathematics is tough but let me tell you
59:31this mathematics topic significant
59:33figures in chemistry is very very easy
59:37and interesting there are certain rules
59:39for significant figures and these rules
59:41will help you to find out the value of
59:43significant figures in certain questions
59:46so there are some rules the very first
59:48rule is that all non-zero digits what
59:50are the non-zero digits from first
59:52towards the ninth one zero is not
59:54included in that right so all non-zero
59:57digits are significant now let us take
59:59an example before example please
1:00:02underline this okay it is very important
1:00:05rule take an example 285 centimeter
1:00:08please let me tell you
1:00:11is 2 a non-zero digit yes ma'am is eight
1:00:15a non-zero digit yes ma'am five is a
1:00:17nonzero digit yes ma'am so how many
1:00:19significant figures are present three
1:00:21significant
1:00:23figures are present
1:00:25you just need to find out the non-zero
1:00:27digits because all non-zero digits are
1:00:30significant so how many significant
1:00:33figures here here are three okay one
1:00:35more example is 0.25
1:00:38so how many significant figures are
1:00:39there i have told you that only non zero
1:00:42digits
1:00:44zero point two five contains zero which
1:00:45is zero digits so this will not be
1:00:48counted hence two and five will only be
1:00:50counted so there are two significant
1:00:52figures hopefully the first rule is
1:00:54clear to you the second rule is zeros
1:00:56preceding to first non-zero digits are
1:00:58not a significant that means if there is
1:01:01any nonzero digit and the zeros are
1:01:03present before that they won't be
1:01:05counted okay so let us take an example
1:01:080.03 so this is the first non-zero digit
1:01:12three is the non-zero digit and if a
1:01:13zero arises before that these zeros are
1:01:15not counted hence it has only one
1:01:17significant figure and coming towards
1:01:20the next case zero point zero zero five
1:01:21two these will not be counted because
1:01:24they are coming before the first
1:01:26non-zero digits hence the answer is only
1:01:28five and two will be counted so answer
1:01:30will be two significant figures so
1:01:32kindly please put it into a box again
1:01:34because this rule is also very very
1:01:36important for you people now coming
1:01:38towards the third rule students what is
1:01:40the third rule i would say the third
1:01:42rule is very simplest zeros between two
1:01:45non-zero digits are significant see
1:01:48zeros before non-zero digit is not
1:01:51significant but between between two
1:01:54non-zero digits is a significant let us
1:01:57take an example this is non-zero digit
1:01:59this is non-zero digit and these are the
1:02:01zeros so these will also be counted and
1:02:04hence there are four significant figures
1:02:06in this case so this rule is also most
1:02:09important so kindly underline them and
1:02:12now we are moving towards the last two
1:02:14rules of the significant figures the
1:02:16last two rules state very easy
1:02:18i have told you the value of 0 before
1:02:20the non-zero digit and in between the
1:02:23non-zero digit now we are talking about
1:02:24the zero value after the
1:02:27non-zero digit and in the case of
1:02:29decimal so let us see zeros at the end
1:02:32or right of the number are significant
1:02:35before and center scene now the right
1:02:37one case or the end one same
1:02:39case the zeros are significant provided
1:02:42they are on the right side of the
1:02:43decimal but they will only be counted
1:02:45when they are at the right side of the
1:02:47decimal now let us take this example
1:02:50will the 0 be counted no because it is
1:02:52before this decimal so it won't be
1:02:54counted
1:02:55this 2 is non-zero yes it is counted and
1:02:57these two zeros are after decimal so
1:02:59they will be counted how many
1:03:01significant figures are there there are
1:03:03total three significant figures here
1:03:05okay but if otherwise the terminal zeros
1:03:07are not significant is there is no
1:03:10decimal point so see
1:03:11if there is no decimal point for example
1:03:14simply 200 is given to you
1:03:16one situation is only 200 you will write
1:03:19no decimal is there you can write it as
1:03:2110 raised to the power 2 we have already
1:03:23done in scientific notations right so
1:03:25these will not be counted and hence only
1:03:28one significant figure but if it is
1:03:30given to you
1:03:31200.0
1:03:33then
1:03:34after non-zero digit this is also
1:03:36counted hence how many significant
1:03:39figures
1:03:40four
1:03:41clear okay now let us take one more
1:03:44example they have taken hundred hundred
1:03:46has only one significant figure why
1:03:48because we can write hundred in terms of
1:03:50one into ten raised to the power two and
1:03:52this is not counted and hence only one
1:03:54significant figure is there but students
1:03:56if i talk about hundred point zero then
1:03:58after decimal one will also be counted
1:04:01and hence they have four significant
1:04:03figures the same thing which i've
1:04:04explained you is written here in both of
1:04:07these situations they have taken 100 as
1:04:09an example 1.0 into 10 raised to the
1:04:11power 2 as an example and 1.00 into 10
1:04:14raised to power 2 as an example now
1:04:16coming towards the last rule that is
1:04:18counting the numbers of objects for
1:04:20example if i say 200 and if i say 200
1:04:23balls so if i say 200 it is referred to
1:04:26as what it is refer to as this situation
1:04:28which is which i have explained you but
1:04:30if i talk about 200 balls i am talking
1:04:33about an object and when i'm talking
1:04:36about an object then it has no limit and
1:04:39hence its significant figures are
1:04:41infinity so in this case you can see
1:04:43that if directly it is given to you two
1:04:45balls 20 eggs so the answer is in finite
1:04:48significant figures because they are
1:04:49exact numbers and we can write these
1:04:51numbers in such way so when there is an
1:04:53object you need to remember this thing
1:04:55students that it has infinite
1:04:57significant figures but if not mentioned
1:04:59then you have to do in the same case
1:05:01okay now let us apply these rules in the
1:05:03following questions the very first
1:05:05question is in front of you
1:05:08okay
1:05:09first thing students as you know
1:05:11questions are there many of the students
1:05:14will skip this portion
1:05:16just for the sake that they are very
1:05:18uncomfortable to do questions
1:05:21let me tell you whenever you face
1:05:23questions you need to tell your mind and
1:05:26your hurt that you can do this you can
1:05:30do this see if you won't do this no one
1:05:32else can do this you know
1:05:34you have to do this for yourself and
1:05:37these questions will only lead you
1:05:39towards the good results and for good
1:05:41results you need to practice so if first
1:05:44time if you are unable to do so don't
1:05:47lose hope just try again second time
1:05:50again they are not they are not uh
1:05:51coming the exact answers no problem
1:05:54don't lose hope try once twice thrice
1:05:57four times five times six times i am
1:06:00assuring you the same that after 10th or
1:06:0315th or 20th or 40th time you will be
1:06:06like give me any question i can do any
1:06:08question because you have understood the
1:06:11concept very deeply by practicing so
1:06:13please practice to achieve your success
1:06:17more practice will leads you towards the
1:06:20success right so first of all i will
1:06:22tell you how to look into the question
1:06:25which approach you should apply towards
1:06:27the question and then we are going to
1:06:29solve we have seen five rules we will
1:06:31consider the rules and we will solve the
1:06:33questions okay with me you are going to
1:06:35do and you are going to achieve it so
1:06:38see the first question is 0.00325
1:06:42here you need to first approach towards
1:06:44this concept that you need to find the
1:06:46non-zero digit what is the non-zero
1:06:48digit non-zero digit is the one in which
1:06:50zero is absent so here this is zero this
1:06:53is zero this is zero hence they want to
1:06:55be counted this is the non-zero digit
1:06:57this is the non-zero digit this is the
1:06:59nonzero digit so
1:07:01all non zero digits are significant and
1:07:04second rule was that before nonzero
1:07:07digit zeros are not counted so before
1:07:10the zeros are not counted hence its
1:07:12answer is 3
1:07:14clear now comes the next case students
1:07:17see this case in this case is
1:07:199.007 we have started this rule if zero
1:07:22comes in between two non-zero uh digits
1:07:25then they are counted this is the first
1:07:27non-zero digit this is the last non-zero
1:07:29digit and hence these zeros will also be
1:07:32counted because they are coming in
1:07:33between hence how many significant
1:07:35figures are present
1:07:37four right very good now coming towards
1:07:40the third question as i told you if
1:07:41object is mentioned to you then you have
1:07:44to write infinity if it is not mentioned
1:07:45you have to directly do this how many
1:07:47non-zero digits three answer is three
1:07:51coming towards the next case fourth case
1:07:53eight point zero zero i told you the
1:07:55after first non-zero digits if any
1:07:57decimal is there and 0 is approaching
1:07:59afterwards then it will be counted so
1:08:01these will be counted after decimal ones
1:08:04answer is 3 now in this case also answer
1:08:08will be one two three and
1:08:11four students now coming towards the
1:08:13next case as i told you that 1 into 10
1:08:16raised to the power 2 we have not
1:08:18considered 10 power 2 hence again 10 to
1:08:20the power 6 will not be included answer
1:08:22will be 1 two and three students now
1:08:25coming towards the next case you can say
1:08:27989 here will be three of the non-zero
1:08:30digits answer will be three the next
1:08:33question all are nonzero digits one two
1:08:36three four 5 answer will be 5. so this
1:08:40was all about our significant figures
1:08:43questions now we are moving towards the
1:08:45round of portion round of motion is very
1:08:47simplest portion we have already done
1:08:49this in you know small classes also what
1:08:52is round this roundup stands for if
1:08:54you're getting some questions in a big
1:08:56decimal number so you do round off to
1:08:58bring it bring it into a small terms for
1:09:01example you have 0.22
1:09:04you have
1:09:060.25 you have zero point term
1:09:103 5
1:09:12you have
1:09:140.27 you have 0.28
1:09:18now what you need to learn is whenever
1:09:21the last digit
1:09:23is more than 5 then the number before
1:09:26them before that will get a number and
1:09:29the last digit will disappear how see in
1:09:31this case 8 is more than 5 the last
1:09:34digit so it will give its one number to
1:09:36to this and will disappear and we'll do
1:09:38round off so what is the answer 0.3
1:09:41because 8 is more than 5. now take this
1:09:43example 0.27 7 is a big number more than
1:09:465 it will give its number to 2 it will
1:09:48be 0.3
1:09:50clear now coming towards the case when
1:09:53the last digit itself is 5 what you will
1:09:56do in that case students when the last
1:09:58digit is 5 but before that there is an
1:10:01odd number here is 3 hence it will be
1:10:04getting one number from it
1:10:07so it will be 0.4
1:10:10but if there is an even number before 5
1:10:13it will not get any point and the answer
1:10:16will be 0.2 because i told you round off
1:10:18only occurs when the number is more than
1:10:205 that's why okay the last case here is
1:10:230.22 students at this case the the last
1:10:26digit is less than 5 hence it will
1:10:28remain same and the answer will be 0.2
1:10:32so this was all about round off
1:10:33situation students you you will be
1:10:35observing some of the questions in
1:10:37addition subtraction multiplication
1:10:38division but in the round a round of
1:10:41case you just need to find out the
1:10:42decimal places in addition and
1:10:44subtraction if you are given some
1:10:46questions just see how many decimal
1:10:49places are present and the answer will
1:10:51contain the same decimal places as of
1:10:54the shortest number but in the case of
1:10:57multiplication or subtraction you have
1:10:58to see the significant figures rule this
1:11:01is not so important just you need to
1:11:03know about the round off i just give you
1:11:05a brief about that situations now we
1:11:07will move forward towards the laws so
1:11:10let us start the law's position and you
1:11:12can note it down and you can practice
1:11:14more questions on this concept
1:11:16law of chemical combination
1:11:19students there are various laws of
1:11:21chemical combination one of them are
1:11:25law of
1:11:27conservation of mass
1:11:36the second one is
1:11:39law of
1:11:42constant proportion
1:11:51students the next one is
1:11:54law of
1:11:56reciprocal
1:12:02proportion
1:12:09the next one is
1:12:12law of
1:12:18multiple proportion
1:12:25and the last one which we will study
1:12:29will be
1:12:29[Music]
1:12:31gail sachs law
1:12:35of
1:12:37volume
1:12:40okay so now let us start from the very
1:12:43first law that is the law of
1:12:45conservation of mass
1:12:46students law of conservation of mass
1:12:49teaches us that mass can neither be
1:12:51created nor be destroyed in any chemical
1:12:54reaction what we observe here is that
1:12:56total mass of reactant is equals to the
1:12:59total mass of product so this is the
1:13:02main important point that we need to
1:13:04remember during the law of conservation
1:13:05of mass that it states us
1:13:09it
1:13:10states us that
1:13:13mass
1:13:16can neither be created
1:13:23nor be destroyed
1:13:27nor be
1:13:29destroyed
1:13:32in any reaction
1:13:37so the most important point
1:13:40that
1:13:41total
1:13:43mass of
1:13:45reactants
1:13:49is equal to
1:13:51the total mass of
1:13:54product it is very simple students okay
1:13:57the total mass of products
1:14:03right
1:14:05kindly put this into a box
1:14:08so that you remember for your further
1:14:11questions now let us apply this formula
1:14:14into our very first question of law of
1:14:16conservation of mass what mass of silver
1:14:19nitrate will react with eight
1:14:235.85 grams of sodium chloride to produce
1:14:2814.35 grams of silver chloride and 8.5
1:14:32grams of sodium nitrate
1:14:35if the law of conservation of mass is
1:14:37true so we need to prove it if it's true
1:14:39or not
1:14:40okay if it's true then we have to follow
1:14:43this and we need to find what we need to
1:14:45find the mass of silver nitrate right
1:14:49okay first of all i'm going to write the
1:14:51given situations which is given to me
1:14:53the given substances and their masses
1:14:55right so it's given that
1:14:57silver nitrate it means agno3
1:15:03it will react with the
1:15:055.85 grams of sodium chloride what is
1:15:08sodium chloride it is nacl
1:15:12to produce now it is producing something
1:15:14that is product is being formed okay
1:15:16then we are going to apply narrow
1:15:18now what is being produced silver
1:15:21chloride that means
1:15:23silver chloride is our agcl
1:15:27and 8.5 grams of sodium nitrate sodium
1:15:32is n a and nitrate will be nano3 so
1:15:35students first of all write this as like
1:15:39this these are our reactants
1:15:42which react together
1:15:44to form
1:15:46products
1:15:47right
1:15:49you all know these things good
1:15:51now students
1:15:52we have been given that we need to find
1:15:55the mass of silver nitrate so let us say
1:15:57that its mass is x grams because
1:16:00everything is taken in grams so we
1:16:03assume that the mass of silver nitrate
1:16:05is x grams it is reacting with nacl
1:16:09whose mass is given to us 5.85 grams so
1:16:12it is 5.85 grams right then it is
1:16:16producing us agcl which is about 14.35
1:16:20grams right
1:16:2114.35 grams moreover it is producing
1:16:25nano 3 which is 8.5 grams so now as you
1:16:29all know according to
1:16:32according to
1:16:36law of
1:16:38conservation of mass
1:16:40according to
1:16:42law of
1:16:43conservation of mass
1:16:47mass can neither be created nor be
1:16:49destroyed hence what do we say that we
1:16:52say
1:16:53total mass of reactant
1:16:59is equal
1:17:00to total
1:17:03mass of
1:17:04product
1:17:06clear students so
1:17:09what will be the general formula general
1:17:11formula will be
1:17:13a gno3 mass
1:17:15plus nacl mass
1:17:18will be equal to the products mass
1:17:21agcl
1:17:22plus nano 3
1:17:25hence we can write that
1:17:27x grams plus
1:17:305.85 grams is equal to
1:17:3414.35 grams plus 8.5 grams okay now this
1:17:39is x plus 5.85 grams is equals to we are
1:17:43going to add these it is going to be
1:17:46students
1:17:51okay
1:17:52right it is going to be
1:17:5515.15 grams
1:17:57let me check this
1:18:0014.35 grams plus 8.5 grams right
1:18:04okay
1:18:05okay now we are going to take this on
1:18:08another side
1:18:09right
1:18:12one second kindly correct this
1:18:15yes
1:18:17just to solve this in a better situation
1:18:19what you can do is if you are getting
1:18:20this at a one decimal place put a zero
1:18:23aside here so that it will be easy for
1:18:25you to calculate now 5 will come here
1:18:29and this is 8 and 8 and 4 is 12
1:18:32and 1 is carried forward so it is 22.85
1:18:36clear okay now we are going to this on
1:18:38this side and our x value will come out
1:18:41so
1:18:42x will come out to be 22.85
1:18:45minus 5.85 which is about 17 gram
1:18:49students so our answer is 17 grams okay
1:18:53clear simplest thing that we need to
1:18:55remember in the law of conservation of
1:18:57mass is total mass of reactant should be
1:18:59equal to total mass of product clear
1:19:01very good now proceeding towards our
1:19:03next law that is the law of constant
1:19:05composition basically law of constant
1:19:08composition defines us that when two
1:19:10elements combine with each other they
1:19:13combine in a simple fixed whole number
1:19:16ratio okay now just take an example if i
1:19:19want water okay i am taking a water from
1:19:23uh you know at a mountain hills i am
1:19:26taking water from river i am taking
1:19:28water from lakes i am taking water from
1:19:31pond the water i am taking from
1:19:33different region
1:19:35is still composed of fixed ratio water
1:19:39is
1:19:40made up of like
1:19:42you know
1:19:44two ratio of the like oxygen there are
1:19:46two parts of hydrogen combining with one
1:19:48parts of oxygen that means if i say what
1:19:50is the ratio of water water ratio will
1:19:53be 2 ratio 16 that means 1 ratio 8 so if
1:19:57i go on a mountain or if i go on a river
1:20:00or if i go in a lake side to get water
1:20:03the ratio will be still the same at any
1:20:06places so this is what this is the law
1:20:08of constant proportion that when two
1:20:10elements the two elements are what
1:20:13hydrogen and oxygen when they combine
1:20:15with each other then the fixed
1:20:18mass or the fixed weight is observed
1:20:21they combine in a fixed ratio of weight
1:20:23or fixed ratio of masses right so we can
1:20:26write that it states that
1:20:31it states us that the elements
1:20:36combined
1:20:39in
1:20:40a
1:20:41fixed
1:20:45ratio by weight
1:20:49clear okay students now coming towards
1:20:52our next law that is the law of multiple
1:20:54proportions
1:20:56okay we are going to do the next law
1:20:59that is
1:21:01law of
1:21:03multiple proportions the best way to
1:21:06deal with laws students kindly
1:21:08understand the concept like in this law
1:21:11what we are observing we are observing
1:21:12about the two elements concept that two
1:21:14elements are combining with each other
1:21:16and they are present in the fixed you
1:21:18know um ratio so whenever you see the
1:21:21questions and there are the three
1:21:23compounds given to you kindly consider
1:21:25that those three compounds contain the
1:21:28same kind of elements so that you can
1:21:30identify that this is the kind of law of
1:21:32constant composition because there will
1:21:33be the same elements then only there
1:21:35will be the fixed ratio of these
1:21:37elements right so when you come across
1:21:39the law of multiple proportion kindly
1:21:40consider this that there will be at
1:21:42least three elements in this situation
1:21:44three different elements in this
1:21:46situation so in the case of law of
1:21:48multiple proportion what we say that
1:21:52when two elements
1:21:57combined with
1:22:01when
1:22:02two elements
1:22:03combined with the fixed you know the
1:22:07fixed mass
1:22:14of the other element
1:22:20they are present in
1:22:29fixed ratio
1:22:31with each other
1:22:39okay to understand this in a very better
1:22:42form i would take an example of law of
1:22:44multiple proportion if i say that
1:22:47of carbon dioxide and carbon monoxide i
1:22:50am taking two examples it's like carbon
1:22:53dioxide and carbon monoxide so what i am
1:22:57observing is this when two elements
1:22:59combine with the fixed mass of other
1:23:01elements like in carbon uh oxygen here
1:23:05one element like if i am taking off
1:23:07carbon carbon is in one part oxygen is
1:23:09in one part while if i take in this
1:23:12carbon is in one part while oxygen has
1:23:14two parts so in this case carbon uh
1:23:16monoxide co we observe that ratio of
1:23:20carbon to oxygen is going to be 12 ratio
1:23:2316 while if i take of co2 what i observe
1:23:26is the ratio is going to be 12 ratio 32
1:23:30by 32 because 1 oxygen has a mass of 16
1:23:32and 2 oxygen will have a mass of 32 so
1:23:35what i observe is one
1:23:37element has a fixed mass the fixed mass
1:23:40one element has the fixed mass while the
1:23:43other elements will be present in fixed
1:23:45ratio with each other hence the oxygen
1:23:48ratio will be 16 ratio 32 or ratio o2
1:23:52will be 16 ratio 32 which is 1 ratio 2
1:23:55so it is a fixed ratio what we i'm what
1:23:58i'm trying to
1:24:00convey to you that whenever you are
1:24:02looking to the law of multiple
1:24:04proportion you will see that there will
1:24:06be elements like
1:24:09in both of these elements one element
1:24:11will have a fixed mass and that will be
1:24:13combining with the other element and
1:24:16that other element will have the fixed
1:24:19ratio with the compound of the other
1:24:21element like in this this oxygen will
1:24:23have fixed ratio with this o2 which is
1:24:25going to be one ratio so this is the
1:24:28example of law of multiple proportion
1:24:30now let us take an example over this um
1:24:322.16 grams of copper metal when treated
1:24:35with nitric acid followed by ignition of
1:24:38the nitrate gave
1:24:402.70 gram of copper oxide in another
1:24:43experiment 1.15 gram of copper oxide
1:24:46upon reduction with hydrogen gave
1:24:490.92 grams of copper show that the above
1:24:52data illustrate the law of definite
1:24:54proportions so students now only we have
1:24:57started the law of constant proportion
1:24:59which are also known as law of definite
1:25:02proportions or definite composition so
1:25:04don't get confused what is law of
1:25:06definite proportions it is still the
1:25:08same now what we are going to do is
1:25:12as i have already told you in this case
1:25:15what it tells us that there will be
1:25:18always a fixed ratio by weight so we are
1:25:20going to find that only now taking the
1:25:22example 2.16 gram of copper metal
1:25:272.16 gram of
1:25:29copper metal you know it is treated with
1:25:32nitric acid followed by ignition of the
1:25:34nitrate which is producing 2.70 gram of
1:25:37copper oxide
1:25:39like 2.16 gram of co is producing
1:25:442.70 gram of
1:25:46copper oxide
1:25:51clear in another experiment what is
1:25:54going to happen students in other
1:25:56experiment zero point do it in other
1:25:59experiment
1:26:001.15 grams of copper oxide upon
1:26:04reduction with hydrogen gave
1:26:060.9 grams of copper
1:26:09kindly consider this in the next one
1:26:13in the next situation experiment what we
1:26:16are seeing that this much copper oxide
1:26:18on reduction with hydrogen is give
1:26:20giving us this much copper so we need to
1:26:23show that this data is following the law
1:26:25of definite proportion how can we
1:26:28explain them we will find the percentage
1:26:30of copper and oxygen in copper oxide so
1:26:33first of all i will find the percentage
1:26:35of cu
1:26:37in copper oxide so what i am going to do
1:26:39the mass of c which is given to me
1:26:40divided by the total mass of the
1:26:42substance that is the copper oxide into
1:26:44100 so it is 2.16 divided by 2.70
1:26:49into it is going to be 100 in this way
1:26:52students we are going to calculate such
1:26:54situation and uh we are going to get 80
1:26:57percent here right when you are going to
1:26:59solve this it is about 80 percent so
1:27:01what we have found that cu percentage in
1:27:05copper oxide in the experiment one i am
1:27:07talking about experiment one in
1:27:09experiment one it is eighty percent so
1:27:12the leftover percentage is of oxygen
1:27:14right so percentage of oxygen
1:27:17in copper oxide in experiment 1 it is
1:27:20going to be 20 right now let us see the
1:27:23next case the experiment 2 case if i
1:27:26talk about experiment 2 students in
1:27:28experiment 2 what it is given is
1:27:300.9 grams of copper is produced
1:27:35from how much copper oxide from 1.15
1:27:39grams of copper oxide
1:27:41so again i'm going to find here the
1:27:44percentage of copper as well as of
1:27:47oxygen so if i found the percentage of
1:27:49copper what i need to do is i need to
1:27:51write the mass of copper upon the mass
1:27:54total mass in 200
1:27:56same 0.92 divided by 1.15 into we'll do
1:28:00100 because we need to find the
1:28:02percentage students so when we calculate
1:28:05its students it is about 80 percent
1:28:10okay it is about 80 percent so the
1:28:13leftover is oxygen hence we can say that
1:28:15percentage of oxygen
1:28:18is
1:28:1920 in experiment two so now tell me as i
1:28:23already told you in the previous case
1:28:24that water in any area in any aspect
1:28:28will have the same you know the same
1:28:30fixed ratio of their masses of their
1:28:33weight wherever you go wherever their
1:28:35elements will be in six fixed ratio so
1:28:38you can see in both of these experiments
1:28:40they have fixed ratio 80 ratio 20 80
1:28:42ratio 20 right so they have fixed ratio
1:28:45whatever maybe the situation is so yes
1:28:47it illustrates the law of definite
1:28:50proportions clear good now students
1:28:53coming towards our next question two
1:28:55oxidation metals contains
1:28:5827.6 percent and
1:29:0130 percent of oxygen respectively
1:29:04if formal of the first oxidation is m3o4
1:29:07find the formula for the second one so
1:29:10it is very simply it has given us that
1:29:12there are two situations in which metal
1:29:15is being reacted with the oxygen and it
1:29:17is forming metal oxides there are two
1:29:19experiments but for the first one we
1:29:21have been given the formula but for the
1:29:23second one the formula is not given so
1:29:25how can we calculate this first of all
1:29:27write the experiment one or the
1:29:29situation one in the experiment one what
1:29:32we observe is the formula has given us
1:29:34which is m three o four
1:29:36right m m3o4 and we have already been
1:29:39given the percentages of oxygen in both
1:29:41the cases like in the first case the
1:29:44case they are saying 27.6 percent of
1:29:46oxygen is present in the first one
1:29:52in first case
1:29:54percentage of oxygen
1:29:58is
1:29:5927.6
1:30:02now students please just uh just tell me
1:30:04that
1:30:05m3o4 and the percentage of oxygen is
1:30:0827.4 we can say that how many oxygen
1:30:12atoms are present here four oxygen atoms
1:30:15are present there but before that tell
1:30:17me the percentage of metal here how can
1:30:19you calculate look whenever such
1:30:22questions arises always consider that
1:30:24the total percentage is 100 hence if
1:30:26oxygen is present 27.6 percent what will
1:30:29be the metal 100 minus 100 minus
1:30:3327.6 percent it is going to be
1:30:4072.4 percent my dear students it is
1:30:42going to be 72.4 percent now we have
1:30:45calculated the percentage in the first
1:30:47case where we have been given the
1:30:48formula right now what we are going to
1:30:51do
1:30:54we are going to do we are going to write
1:30:56the second case right in the second case
1:30:58in the second experiment what
1:30:59information i have got i don't have the
1:31:01formula but i know that percentage of
1:31:03oxygen is 30 percent right so the
1:31:05percentage of oxygen in the second case
1:31:08is 30
1:31:09so the other portion obviously will be
1:31:11the metal right so i would say that
1:31:13percentage of metal would be
1:31:16100 minus 30 hence it will be 70 percent
1:31:21now in both these cases we have already
1:31:23calculated the percentages now what will
1:31:25be the other
1:31:26next step in the next step we will
1:31:29observe that here four parts of oxygen
1:31:32you know have percentage 27.6
1:31:36right so how much i would say oxygen
1:31:39will be present for the next case we
1:31:41need to calculate it
1:31:43it is little bit of mathematics you can
1:31:45consider but don't get confused by step
1:31:48i am going to explain you first of all
1:31:50i'm going to explain you about the metal
1:31:52m
1:31:53see three parts these are three
1:31:56three parts of metal
1:32:01gives
1:32:0372.4 percent of metal right you know
1:32:06this thing now if i say one part of
1:32:09metal will give
1:32:12how much
1:32:13one part of metal will give you how much
1:32:16this you need to carry this you need to
1:32:19calculate so
1:32:22in other words we need to calculate this
1:32:24but it has already told us that the
1:32:27percentage of metal is 70 percent in the
1:32:29another case so we have been given 70
1:32:32percent is already present we don't know
1:32:35how many parts
1:32:37of this metal will be formed
1:32:40we don't know this situation we just
1:32:42only know that three parts are giving us
1:32:4572.4 percent so how many parts will be
1:32:49formed when 70 percent is there in short
1:32:52reverse the situation to do this
1:32:53question it will be easy for you i would
1:32:56say 72 percent
1:32:58is present in three parts
1:33:01so in seventy percent how many parts
1:33:04will be present it will be three by
1:33:06seventy two point four into 70.
1:33:09like this you can do the question first
1:33:11of all write the situation understand
1:33:12the situation what they are going to
1:33:14tell you so whenever you will solve the
1:33:16students it will be
1:33:17um
1:33:18approximately
1:33:202.90 something
1:33:22it will be approximately this much now
1:33:24what is this these are the parts of
1:33:26metal that arises here are three parts
1:33:29now you have uh already found how many
1:33:31parts are there how many parts are there
1:33:34of metal
1:33:35two point nine zero parts of metal in
1:33:37case two
1:33:38similarly we are going to calculate for
1:33:40the oxygen part now see again i am
1:33:43telling you here the percentage is given
1:33:45twenty seven point six percent twenty
1:33:46seven point six percent gives four parts
1:33:49so how many will thirty percent give
1:33:51same thing arises i'm saying twenty
1:33:53seven point six percent gives us
1:33:57four parts right
1:33:58now i am telling you seventy percent
1:34:01will give four by twenty 27.6 into 70 my
1:34:06dear students it will be about four
1:34:09point
1:34:13three five
1:34:15four point three five parts of
1:34:18oxygen okay now we have already
1:34:21calculated the parts of oxygen as well
1:34:23as the parts of
1:34:25metal now as you know we do the ratio
1:34:27forms so we are going to write it here
1:34:31metal ratio oxygen for the second case
1:34:34now firstly you need to write for the
1:34:36metal for the metal we have calculated
1:34:392.90 for oxygen we have calculated
1:34:43four point three five minus students it
1:34:45is going to be near about one ratio i
1:34:49would say one point five aprox one point
1:34:52yes so one ratio one point five so in
1:34:56short whenever we get it in the decimals
1:34:58we try to convert it into a fixed number
1:35:01so that it is easy for us so to convert
1:35:031.5 into a number i would say just
1:35:06multiply it with 2 that would be easy
1:35:08for you so if you multiply with 2 it
1:35:10will be 2 ratio
1:35:123 students hence what will be the
1:35:14formula
1:35:15m
1:35:16uh base two
1:35:18oxygen base three so the formula that is
1:35:21obtained for the second part is m two o
1:35:23three it is clear to you it is very
1:35:24simplest question but it the students
1:35:26get confused and disconfused in this
1:35:28question like in this part of the
1:35:30question so you need to understand first
1:35:33write situation in such a way and find
1:35:35out what a new need to calculate you
1:35:37need to calculate the parts so kindly
1:35:39convert this in this form it is a simple
1:35:41basic mathematics so that it will be
1:35:43easy for you to calculate the formula
1:35:45and formulate it right so through this
1:35:48question we can elaborate and calculate
1:35:51different questions now moving forward
1:35:53towards the next law that is the law of
1:35:55reciprocal proportions students law of
1:35:58reciprocal proportions states uh between
1:36:00the three elements like
1:36:03if i say there is carbon there is
1:36:05hydrogen there is oxygen what happens is
1:36:08carbon forms bond with hydrogen and
1:36:10forms a compound which is ch4 then
1:36:16i see a case where carb uh carbon forms
1:36:19bond with oxygen and the compound that
1:36:22is formed formed is co2 in the next case
1:36:25students hydrogen oxygen also combines
1:36:28with each other forming water so what is
1:36:30the law of reciprocal proportion it
1:36:32stays states us that when two elements
1:36:35consider any two elements for example i
1:36:37am considering carbon and oxygen when
1:36:40these two elements combine with the
1:36:42fixed mass of the third element
1:36:44of the third element means hydrogen so
1:36:47first see what i am speaking i am saying
1:36:49when carbon
1:36:51combines with hydrogen
1:36:53it forms a ch4 what will be their ratio
1:36:57students in ch4 it will be 12 ratio 4
1:37:01right now
1:37:03when oxygen combines with hydrogen
1:37:06students oxygen combines with hydrogens
1:37:09forming water what will be their ratio
1:37:1216 ratio 2 right we need to make one of
1:37:17the elements ratio constant so i am
1:37:19making constant the ratio of hydrogen
1:37:21because i am considering these two
1:37:23formulas you can consider any of the
1:37:24formulas by taking one as a constant
1:37:27fixed ratio i am taking hydrogen's fixed
1:37:30ratio so in this case hydrogen is four
1:37:32parts in this case hydrogen is two parts
1:37:33so what i'm going to do is i'm going to
1:37:35multiply this with two now what i'm
1:37:37getting is oxygen ratio hydrogen that
1:37:39means when it is multiplied with 2 it is
1:37:4232 ratio 4 students clear now the new
1:37:45ratio is 12 ratio 4 for ch4 and for
1:37:48water after multiplying it with 2 we are
1:37:50getting 32 ratio 4 what happens is the
1:37:53ratio of hydrogen is now fixed so what
1:37:56we are observing we are observing that
1:37:58if the ratio of
1:38:00if
1:38:01ratio of
1:38:03hydrogen is fixed
1:38:06then
1:38:08then
1:38:09carbon oxygen ratio will be
1:38:1312 ratio 32 clear students it is
1:38:17simplest or i could say or i could say
1:38:21if i divide it on the table of 2 i would
1:38:24say
1:38:266 ratio
1:38:28it's
1:38:3016 right
1:38:32it is
1:38:336 ratio 16
1:38:34uh if i divide on the table of 2 or if
1:38:37you further divide it on 2 it will come
1:38:39out to be
1:38:403 ratio 8 clear simplest okay now
1:38:45students i have found that after fixing
1:38:48the ratio of hydrogen the ratio of
1:38:51carbon and oxygen is 3 ratio 8. now
1:38:54individually what will be the ratio let
1:38:56us calculate that individually if i
1:38:59consider both of these cases when carbon
1:39:01directly combines with oxygen when
1:39:04carbon
1:39:06directly combines
1:39:09directly combines with oxygen students
1:39:15what we get what the what is the ratio
1:39:17we get
1:39:1812 ratio 16
1:39:21now i will again divide it on the table
1:39:24of 2 what i'm getting 6
1:39:26ratio i am getting 8 what i am getting
1:39:29here is 6 ratio 8 but please do remember
1:39:33the compound hair formed is carbon
1:39:34dioxide not carbon monoxide so here it
1:39:38will be 16 into 2 hence 2 into 8 or i
1:39:41would say again if i cut it on the table
1:39:43of 3 it is 3 ratio 8 on the table of 2
1:39:47sorry it will be 3 ratio it here the
1:39:50formula is a co2 not co so you have to
1:39:54remember in your minds the formulas
1:39:56first formula is ch4 accordingly you
1:39:58will write the ratio second formula is
1:40:00water accordingly you will write the
1:40:02ratio third formula is co2 accordingly
1:40:05you will write the ratio so what is the
1:40:07thing that is we are observing here we
1:40:09are observing that when we are
1:40:11considering the two elements in keeping
1:40:13the ratio of the uh third element fixed
1:40:15we are getting both of these ratio is
1:40:17three ratio it which is equal
1:40:19to the situation when they both combined
1:40:22directly hence this is law of reciprocal
1:40:25proportions hopefully it is clear to you
1:40:28now moving forward towards of a question
1:40:30hydrogen combines with
1:40:33chlorine to form hcl claire i'm going to
1:40:36write the situation it will be easy for
1:40:37you right hydrogen what it does it
1:40:40combines with chlorine right to form
1:40:44hcl okay
1:40:46the next one is sodium combines with
1:40:50hydrogen again to form nh
1:40:55okay
1:40:56now
1:40:57it says if sodium and chlorine also
1:41:00combine with each other
1:41:02sodium and chlorine also combined with
1:41:06each other students they will do so in
1:41:09the ratio of their masses we need to
1:41:10calculate the ratio of the masses now we
1:41:13have already done the situation see
1:41:16we have already calculated the situation
1:41:20see hydrogen
1:41:22chlorine and i would say sodium
1:41:25when hydrogen combines with chlorine it
1:41:27is forming hcl when chlorine combines
1:41:30with the sodium it is forming nacl when
1:41:33hydrogen combines with na it is forming
1:41:36nh so
1:41:37look into the situation like we did
1:41:39previously first of all in the case of
1:41:41hcl
1:41:43the ratio will be 1 ratio 35.5 we are
1:41:46always taking the masses their weight
1:41:48right in the case of nah
1:41:51again
1:41:53sodium ratio hydrogen na is 23 while
1:41:56hydrogen is 1. so what we are observing
1:41:59here the hydrogen uh mass is already
1:42:02what it is already fixed right if it is
1:42:05already fixed my dear students so what
1:42:07we observe in this situation that n a
1:42:10and c l will have what n a and cl
1:42:16will have 20 to 3 ratio 35.5
1:42:20clear because hydrogen masses are now
1:42:23constant in both of the cases so the
1:42:25leftover is these both and these are 23
1:42:27ratio 35.5 if we did individually also
1:42:30then also it will be coming out 23 ratio
1:42:3335.5 now you will say that i mean these
1:42:36options a and b are almost similar how
1:42:39to identify please look at the formula
1:42:41the formula is nh so first of all you
1:42:44have to write the ratio the mass of na
1:42:47then hydrogen na is 23 then uh sorry nah
1:42:52and nacl and hcl it is like this now and
1:42:56it is being asking about nacl so first
1:42:58of all we will write about na then we
1:43:00will write about cl so n is 23 then is
1:43:0335.5 so which option will be correct a
1:43:06option will be correct answer will be
1:43:08option number a
1:43:10clear okay now proceeding towards the
1:43:13next questions my dear students which of
1:43:15the following combination illustrate the
1:43:18law of a reciprocal proportion we have
1:43:21already understood in the law of
1:43:22reciprocal proportion we at least need
1:43:24three elements that is being contact
1:43:27with each other right so in the first
1:43:29case only nitro ocean and oxygen in the
1:43:31second also compound nitrogen oxygen in
1:43:34third also so this is not possible at
1:43:36all because we need at least three
1:43:37elements different in second b option
1:43:40again phosphorous hydrogen phosphorus
1:43:41oxygen phosphorus sulfur see
1:43:44likewise i gave you the previous case
1:43:47what we did was hydrogen chlorine and na
1:43:49the three had connection with each other
1:43:52but here what you see phosphorus
1:43:54hydrogen oxygen sulfur it is not
1:43:57possible it is not possible four
1:43:59elements no so it is not considered now
1:44:01the c option carbon is present
1:44:04sulfur is present oxygen is present
1:44:06carbon is forming bond with sulfur which
1:44:08is cs2 carbon is
1:44:11bonding with oxygen forming co2 and
1:44:14sulfur is forming bond with oxygen as
1:44:16so2 so you can directly see that it
1:44:18follows the law of reciprocal proportion
1:44:21hence the answer c will be our correct
1:44:24one clear very good students now we are
1:44:27proceeding towards the last law that is
1:44:29the gay lussac's law of combining
1:44:31volumes what does it states us it states
1:44:33us that according to this law when the
1:44:36gas sees please consider this when gas
1:44:38is reacted together they always do so in
1:44:42the volume which bear a simple ratio
1:44:45a simple ratio to one another and to the
1:44:48volume of products for example if i say
1:44:51h2 reacts with cl2 to form hcl students
1:44:55first of all balance it out here are two
1:44:57elements of hydrogen here are two
1:44:59elements of chlorine here are only one
1:45:01and one elements of hydrogen and
1:45:02chlorine so what you'll do is you'll put
1:45:04a twice number here so that it is
1:45:06balanced out right so what we say that
1:45:09here it is one one
1:45:11right so if i say that if here it is one
1:45:13part i would say if it is gaseous so
1:45:16here one volume of h2 reacts with one
1:45:19volume of h2 to a cl2 to give two volume
1:45:23of hcl how i can say that on the basis
1:45:26of the stoichiometry on the basis of
1:45:28these numbers which are present in front
1:45:30of them like one h2 is combining with
1:45:32one cl2 to give two hcl that means one
1:45:34volume of uh h2 is combining with one
1:45:37volume of cl2 to give two volume of hcl
1:45:39so in which ratio are they they are in
1:45:41one ratio one ratio to they bear a
1:45:45simple fixed ratio so this is the law
1:45:48according to this law when gases
1:45:50react with each other they do so in
1:45:52volumes in volumes right which bear a
1:45:56simple ratio to one another to one
1:45:58another they bear a simple ratio but
1:46:00also to the volume of product also to
1:46:03the volume of product if these are also
1:46:05gases provided all the measurement of
1:46:07the volumes are done under similar
1:46:09conditions of temperature and pressure
1:46:11they are done under the similar
1:46:12conditions of temperature and pressure
1:46:14so this was all about the gay lussac's
1:46:15law you need to just put the
1:46:18stoichiometric correct that's it if you
1:46:20will able to do this balancing out then
1:46:23you can find this ratio very simply
1:46:25first balance out then you can find the
1:46:27ratio it is as simple as it is right
1:46:30students and now comes the question part
1:46:33just look at the question if law of
1:46:35conservation of mass was to hold it true
1:46:38now what was law of conservation of mass
1:46:39when total mass of reactant is equals to
1:46:41the total mass of product this was the
1:46:44law of conservation of mass then 20.8
1:46:47grams of bscl2
1:46:50bacl2 how many
1:46:5220.8 grams
1:46:54reaction with
1:46:569.8 grams of h2so4
1:47:00will produce
1:47:017.3 grams of hcl let us write it down
1:47:07okay it is giving us 7.3 grams of hcl
1:47:11and
1:47:12baso4 equals two we need to find out the
1:47:16mass of bs so force it is as simple
1:47:19what was the formula of law of
1:47:20conservation of mass
1:47:22total
1:47:24mass of reactant
1:47:28is equal to
1:47:31total
1:47:34mass of product
1:47:36students now the total mass of reactant
1:47:39will be 20.8 plus 9.8 is equals to 7.3
1:47:44plus x right first of all we'll add them
1:47:49it is 30.6 students it is going to be
1:47:5430.6 let me just do this again yeah so
1:47:58is equals to 7.3 plus x now this will
1:48:01come here and subtract 30.6 minus 7.3 is
1:48:05equals to x which is going to be
1:48:083
1:48:093 so it is going to be
1:48:1123.3 grams my dear students now let us
1:48:14see the options hence b option is our
1:48:17correct one clear so the law of
1:48:19conservation of mass states is that
1:48:21total mass of reactant is equals to
1:48:22total mass of product so here students
1:48:26comes the dalton atomic theory after the
1:48:28loss dalton atomic theory states us
1:48:31about atom
1:48:32basically dalton atomic theory tells us
1:48:34that matter comprises of very small
1:48:38extremely small indivisible particles
1:48:41that cannot be divided students which
1:48:43are known as atoms which are known as
1:48:46atoms and these atoms these atom
1:48:49students of a given element are
1:48:51identical in all aspects in which which
1:48:54expects their identical that they have
1:48:57same size they have same shape they have
1:49:00same mass they have same chemical
1:49:03properties the next thing is that atoms
1:49:05of different elements are different in
1:49:08all respects see atoms of same element
1:49:11will have all the same properties but
1:49:12atoms of different elements will exhibit
1:49:14different properties for example they
1:49:17possesses different say shapes sizes
1:49:20mass in all of these right then it comes
1:49:23that atoms of different elements may
1:49:25combine with each other they can combine
1:49:27with each other
1:49:28in fixed simple or whole number ratio to
1:49:31form compounds clear they can form
1:49:33compounds next thing is atoms can
1:49:35neither be created nor be destroyed in a
1:49:37chemical reaction students but there
1:49:39were the limitations of dalton atomic
1:49:41theory which couldn't explain us about
1:49:44isotopes you know the second thing was
1:49:46that they were they weren't unable to
1:49:48explain about how they were able to form
1:49:52compounds how different elements could
1:49:54combine and form compounds so these were
1:49:56the limitations of dalton atomic theory
1:49:59right so now we are proceeding towards
1:50:01our next topic that is going to be
1:50:03atomic and molecular mass but before
1:50:05atomic and molecular mass we need to
1:50:07understand about atom which we have
1:50:09already done in dalton atomic theory
1:50:11that atom is the smallest indivisible
1:50:13particle and it is uh responsible for
1:50:16any chemical reaction
1:50:18right but what we see is that when we go
1:50:20into market what is the need you know of
1:50:23mole concept of bringing this chapter to
1:50:26you the main reason is that there are
1:50:28many quantities which we can measure but
1:50:31there are some quantities which we
1:50:32cannot measure to measure them we have
1:50:35introduced mole concept how i am saying
1:50:37that if i go to market to buy some
1:50:40fruits or bananas i say that i need one
1:50:43dozen bananas one doesn't stand so 12
1:50:46bananas right and um then
1:50:49the next day i go and i ask for five
1:50:51dozen so how many bananas i'm going to
1:50:53get 60 bananas i'm going to get so what
1:50:55i've done for bananas i have set an
1:50:57equation i've set a set of standard like
1:51:00dozens right so i can calculate that in
1:51:02terms of fruits and vegetables but when
1:51:04i come across chemistry there are some
1:51:07quantities there are some atoms whose
1:51:10amount of substance is not been
1:51:12calculated in terms of dozens or kgs or
1:51:15like this no so we need to introduce a
1:51:17term and hence we introduced atom
1:51:20so these atom these atomic mass helped
1:51:24us to calculate the amount of substance
1:51:26in chemistry field in chemical reactions
1:51:29so whenever we could correlate this with
1:51:31the daily life situation for daily life
1:51:33we have
1:51:34referred a standard similarly for
1:51:37chemical reactions we have kept a
1:51:39standard that is c12 carbon as our
1:51:42standard reference with the help of that
1:51:44we are going to calculate all the masses
1:51:47like with the help of carbon atom we can
1:51:49calculate other atoms element masses and
1:51:52through them we can calculate atomic
1:51:54mass and molecular mass so what is going
1:51:56to be the atomic mass runes atomic mass
1:51:59of an element is number of times an atom
1:52:01of that element is heavier than an atom
1:52:04of carbon taken as 12. why we are
1:52:07considering here carbon here because as
1:52:09in the case of vegetables we took dozen
1:52:11as in reference in the case of chemistry
1:52:13we took carbon as a reference that's why
1:52:17then the definition which is reported to
1:52:19you is this one that one atomic mass
1:52:21unit which is known as amu
1:52:24it is equal to 1 by 12
1:52:27of the mass of an atom of carbon 12
1:52:30isotope
1:52:31clear students so now we are going to
1:52:34start mole concept in detail considering
1:52:38each and every situation when volume is
1:52:41given to us when molecular mass is given
1:52:43to us so now let us start towards the
1:52:46mole concept so yes students as you all
1:52:49know we are doing mole concept and mole
1:52:51concept first we need to understand what
1:52:53is a mole basically a mole of a
1:52:56substance is
1:52:586.022 into 10 raised to the power 23
1:53:02particles
1:53:03which is referred to as avogadro's
1:53:05number so i would define first of all
1:53:07mole for you what is mole
1:53:10mole is defined as
1:53:16mole is basically defined as one mole of
1:53:19a substance
1:53:25is
1:53:26equal to
1:53:29equal to
1:53:316.022
1:53:34into 10 raised to the power 23 particles
1:53:37now students i am talking about
1:53:39particles here particles are of various
1:53:41kinds it can be electrons it can be
1:53:44protons it can be neutrons it can be
1:53:47atoms it can be ions it can be elements
1:53:51everything lies in these you know
1:53:54particles can be any of them so in short
1:53:57i can say that
1:53:59one mole
1:54:01is equals to
1:54:026.022 into 10 raised to the power 23
1:54:05particles
1:54:08here you can add molecules also instead
1:54:10of elements okay so one mole is equals
1:54:13to 6.02 into 10 power 23 particles
1:54:16particles i've already told you these
1:54:18can be these thing and this value
1:54:20students is known as evo gadro
1:54:24number
1:54:26okay now what is the use of this
1:54:29evocation number we will use this in
1:54:31formulas and it is going to be very
1:54:32interesting and very easy topic okay so
1:54:35now let us proceed towards our next
1:54:38fundamental particles which i already
1:54:39told you about electron protons and
1:54:41neutrons now the particle electron has
1:54:44mass 9.1 into 10 raised to the power
1:54:46minus 31 kg and its charge is minus 1.6
1:54:50into 10 raised to the power minus 19 as
1:54:53it is electrons so negative charge comes
1:54:55in front of this then protons come which
1:54:57are positively charged particles proton
1:55:00basically is represented by small p and
1:55:02its mass is 1.67 into 10 to the power
1:55:05minus 27 kg and its charge is 1.6 into
1:55:0910 raised to power minus 19. while if i
1:55:11talk about neutron then the mass is 1.67
1:55:14into 10 raised to the power minus 27 kg
1:55:16and its charge is neutral so these are
1:55:19the fundamental particles which we
1:55:21should know right okay now one of the
1:55:23question that you they can ask you is
1:55:25calculate charge on one mole of electron
1:55:29basically students to calculate charge
1:55:32on one mole of electron
1:55:36charge on
1:55:38one mole of electron we need to
1:55:40calculate it by
1:55:42number of electrons into
1:55:44charge on one electron
1:55:49see we have to calculate charge of one
1:55:52mole of electron right and for that the
1:55:54formula is number of electron total
1:55:56number of electron into charge of only
1:55:58single electron will help you to
1:55:59calculate the charge on one mole as you
1:56:02all know one mole is basically one mole
1:56:04contains how many electrons we have just
1:56:06done right now one mole contains this
1:56:09kinds of particles particles can be
1:56:10electron that means one mole have this
1:56:13much of electrons right so now here how
1:56:16many electrons are present
1:56:196.022 into 10 raised to the power 23
1:56:22which is avogadro number and it is
1:56:23represented by na clear into now comes
1:56:27the charge on one electron we have just
1:56:29and right now charge is 1.6 into 10
1:56:31raised to power minus 19. you don't have
1:56:33to put this negative sign when you are
1:56:35calculating something it is just to
1:56:36represent that electrons are negatively
1:56:38charged so now we will put the same
1:56:40value 1.6 into 10 raised to the power
1:56:43minus 19 right so when we will solve it
1:56:47it will come 965 double zero coulomb
1:56:51it comes in coulomb you can convert it
1:56:54into faraday also which is equals to one
1:56:56faraday
1:56:57so by this way we can calculate the
1:56:59charge on one mole of electron for
1:57:01example they ask you on two mole of
1:57:02electron then you can multiply it
1:57:04further so these kinds of question can
1:57:06be asked on fundamental particles it is
1:57:08the basic thing that you can get to know
1:57:09right now students i am coming across
1:57:12the most important and interesting part
1:57:14of the chapter which is atomic mass
1:57:16basically what is atomic mass atomic
1:57:18mass is the mass of one moles of atom
1:57:20and there are certain
1:57:22atomic mass that you need to learn for
1:57:24example the atomic mass that you need to
1:57:27be learned you have to learn it in grams
1:57:28basically gram atomic mass is referred
1:57:31to as when we represent mass in terms of
1:57:34grams when atomic mass of atom is
1:57:36represented in grams it is known as gram
1:57:38atomic mass for example hydrogen has
1:57:41gram atomic mass 1. this you need to
1:57:43learn the table which i have drawn here
1:57:45other than this only one two three four
1:57:47would be there that you could recall and
1:57:49learn right but these are important for
1:57:51you to learn to calculate molecular mass
1:57:53okay hydrogen has gram atomic mass one
1:57:56helium has gram atomic mass four lithium
1:57:58has ground atomic mass seven beryllium
1:58:01has gram atomic
1:58:03mass nine then we have further boron
1:58:06which is gram atomic mass 11 then we
1:58:08have carbon as 12 nitrogen as 14 oxygen
1:58:12as 16
1:58:13fluorine has 19 neon as 20 sodium has 23
1:58:18magnesium has 24 aluminum as 27 and you
1:58:22can all they can also ask you about
1:58:23chlorine which is about
1:58:2535.5
1:58:27okay this is a little bit different so
1:58:29they can ask you the same also so you
1:58:31need to learn these elements and their
1:58:33gram atomic mass practice will make you
1:58:35perfect when you will apply them on
1:58:37questions
1:58:38right great students now you need to
1:58:40calculate the atomicity or molecularity
1:58:43what i mean by atomicity or molecularity
1:58:45it tells us how many atoms are present
1:58:47in a molecule so for the first case it
1:58:50is given that c12 h22 o11 how many atoms
1:58:54are present in this case see how many
1:58:56carbon are there 12 how many hydrogen
1:58:59are there 22 how many oxygen are there
1:59:0111 so we'll add them it is
1:59:06it is 45 that means 45 atoms are present
1:59:09in this case now taking this case how
1:59:12many potassium are there two how many
1:59:14sulfur one how many oxygen four how many
1:59:17aluminium two how many sulfur it is into
1:59:20three
1:59:20plus how many oxygen it is four into
1:59:23three that is twelve how many uh
1:59:25hydrogen again twenty four into two it
1:59:28is going to be forty eight plus how many
1:59:30oxygen it is 24 so students by this way
1:59:33when you will calculate it it is about 3
1:59:36plus 4 7 7 plus 9 9 plus 3 15. it is
1:59:40going to be about 96 right clear so
1:59:44basically it tells us about the number
1:59:46of atoms present in a molecule right so
1:59:49we need to calculate all those elements
1:59:51the number of elements right now
1:59:53students the next thing that comes is
1:59:56the molecular mass molecular mass is the
1:59:58mass of one molecule how do we need to
2:00:00calculate it for this you need to learn
2:00:03the table which i already told you you
2:00:05need to individually learn the elements
2:00:07gram atomic mass so that they can you
2:00:09can apply here and you can solve it now
2:00:11let us see for the case of ch4 how many
2:00:13carbons are there one so 1 into carbon's
2:00:16mass plus how many hydrogen is therefore
2:00:194 into hydrogen mass by this way you can
2:00:22calculate what was the mass of carbon in
2:00:24the table it was 12 you can go back and
2:00:26check so 1 into 12 plus what was for
2:00:29hydrogen 4 into 1 so it comes out to be
2:00:3216 so 16 is the molecular mass now
2:00:35coming towards co2
2:00:38again how many carbon are there one so
2:00:40carbon into one that means mass of
2:00:43carbon into one because one ah carbon
2:00:45atom is there two oxygen atoms are there
2:00:48two into oxygen atoms mass so carbon
2:00:51atoms mass is twelve twelve into one
2:00:52plus oxygen mass is 16 16 into 2 so it
2:00:56comes out to be students 44.
2:00:59these table if you have learned then
2:01:01only you can calculate the molecular
2:01:02masses otherwise it will be quite
2:01:04difficult for you now coming towards
2:01:06kmno4 now if i talk about kmno4 students
2:01:10then we need to write the masses for
2:01:13potassium as well as mn right so the
2:01:16mass for
2:01:19one second the mass for potassium is 1
2:01:22into potassium's mass plus 1 into mn's
2:01:25mass
2:01:28then plus 4 into oxygen's mass right
2:01:31potassium mass is 39 right
2:01:34so 39 plus now comes for mn mass it is
2:01:38about 55
2:01:40then comes
2:01:42for oxygen that is 4 into 16 so it is 39
2:01:46plus 55 plus 64. which comes out to be
2:01:5394 plus 54 it is near about 158 students
2:01:57clear now if i talk about c 2 h 5 o h
2:02:00how many carbons are there 2 2 into
2:02:03carbon's mass plus how many hydrogens 6
2:02:05into hydrogen's mass plus 1 into oxygen
2:02:08mass
2:02:09carbon mass is 12 so 2 into 12 plus 6
2:02:12into 1 plus 1 into 16 if we solve it out
2:02:1624 plus 6 plus 16 how much students it
2:02:20is 46 clear now
2:02:24okay coming towards a c4h10
2:02:28so more more practice will make you more
2:02:30clear so first of all learn the gram
2:02:33atomic mass individually for the atoms
2:02:35then apply it here it will be coming in
2:02:37seconds clear
2:02:39c4h10
2:02:40how many carbon are there four four into
2:02:43carbons mass plus 1 into hydrogens
2:02:46number right one is the mass of hydrogen
2:02:49so if we solve it it is 48 plus 10 that
2:02:52is 58 now comes the last part that is
2:02:56urea here
2:02:57okay nitrogen mass is 14 and here are
2:03:00two nitrogens so 14 into two plus there
2:03:03are total four hydrogens and its mass is
2:03:05one so directly i'm writing four carbon
2:03:08is one so i'm writing 12 and oxygen is
2:03:10one so i'm writing
2:03:1216 so it comes out to be 60 so it is the
2:03:15molecular mass for urea clear so by this
2:03:18way you can calculate the molecular
2:03:20masses now students coming towards the
2:03:23formula for the mole concept that is
2:03:24very very important there are three
2:03:27basic formulas that you need to apply
2:03:28number one is number of moles are
2:03:30represented by small n
2:03:33it is number of moles
2:03:36which is equal to
2:03:38given mass
2:03:41upon molecular mass
2:03:43the very first formula molecular mass
2:03:45you have calculated just right now given
2:03:47mass will be given to you in question
2:03:48right this is the first formula the very
2:03:51next formula which is important for you
2:03:53is number of moles
2:03:54in terms of volumes will
2:03:57will be 20 it will be
2:04:00volume of
2:04:02substance
2:04:04it will be a volume of substance divided
2:04:07by 22.4
2:04:10so this is in the case of when you have
2:04:11given the volume you can apply directly
2:04:13this formula and the very last formula
2:04:15that you use for number of moles
2:04:21for the case of molecules or atoms so
2:04:24number of molecules
2:04:27or atoms
2:04:30divided by
2:04:32avogadro's number which is 6.022 into 10
2:04:35raised to the power 23 so these are the
2:04:37three basic formulas that you are going
2:04:39to apply for the questions in the case
2:04:40of mole concept if you have to find the
2:04:42number of moles you can directly use
2:04:44these formulas according to the given
2:04:46situation if you are given given mass
2:04:47then you can calculate this if you
2:04:50want to calculate the volume you're
2:04:51given number of moles you can use this
2:04:53formula if you want to calculate number
2:04:55of moles and you are given number of
2:04:56molecules you can use this formula now
2:04:59students one question is there calculate
2:05:01mass of 2.5 mole of cl2 now number of
2:05:04moles here it is given 2.5 moles
2:05:07right and we have to calculate the mass
2:05:09of cl2 that means given mass of cl2 is
2:05:13missing
2:05:14so we can calculate molecular mass we
2:05:16have just learned that cl2 i have told
2:05:19you how many cl2 into cl's mass cl mass
2:05:23is
2:05:2435.5
2:05:26right so if i solve it it is 71 students
2:05:29now after this number of moles formula
2:05:32is number of moles is equals to given
2:05:34mass upon molecular mass now number of
2:05:36moles is 2.5 given mass we need to
2:05:39calculate and molecular mass students is
2:05:4171. now when we cross multiply it it
2:05:44comes out to be
2:05:48177.5
2:05:50clear
2:05:51so let me calculate it in detail so that
2:05:53it is easy let us check first i guess
2:06:005 two
2:06:06yes
2:06:07it is the same 177.5
2:06:10clear okay now it will come grams
2:06:12because we are taking it in grams right
2:06:15okay students now let us see the flow
2:06:17chart flow chart here is to represent
2:06:20the three basic formulas how three basic
2:06:22formulas can be correlated with each
2:06:24other see if you are given given mass
2:06:31and you are asked to calculate number of
2:06:33molecules you can directly use this
2:06:36formula
2:06:40to calculate
2:06:42how i have used this formula okay i'll
2:06:45tell you about the slide see number of
2:06:47moles is equal to given by super
2:06:48molecular mass also number of moles is
2:06:50equal to number of molecules divided by
2:06:52avogadro number so if left hand side is
2:06:54equal we can put the right hand side
2:06:56equal so this is the core relationship
2:06:58of these formulas which we can apply
2:07:00another with the volume also you can do
2:07:02the same given mass upon molecular mass
2:07:04is equals to volume of
2:07:07substance
2:07:10by
2:07:1322.4
2:07:14right students so this you can use as a
2:07:17correlation of the three basic formulas
2:07:19of mole concept now students next comes
2:07:22our question which has the maximum
2:07:24number of atoms now let us see that the
2:07:26first case is 24 grams of carbon in the
2:07:29case of 24 grams of carbon how many
2:07:32number of atoms are present now
2:07:35see what is the formula of number of
2:07:37moles number of moles is given mass upon
2:07:39molecular mass and also number of moles
2:07:42is equals to number of molecules or
2:07:44number of atoms
2:07:46divided by avogadro number that is
2:07:496.022 into 10 raised to the power 23 so
2:07:51here given mass is 24 molecular mass of
2:07:54carbon is 12 is equals to we need to
2:07:56calculate number of atoms
2:07:58do not write students evocative number
2:08:01in the full form like six point zero two
2:08:03two into it will be very complex so just
2:08:06write it as n a because we need to
2:08:08compare it right so
2:08:10when we divide it it is 2 it is 2 n a so
2:08:13for the first case number of atoms that
2:08:15came out is 2n a now for the next case
2:08:20now for this case it is given mass is 56
2:08:23molecular mass is also 56
2:08:25into na will be equal to number of atoms
2:08:30hence it comes out to be
2:08:321na right okay now the third is
2:08:35aluminium 27 grams of aluminium given
2:08:38mass is 27 and aluminium mass is also 27
2:08:42right and when we'll cross multiply
2:08:44according to this formula into na it
2:08:46will give us number of atoms
2:08:50hence students it's come out to be 1 n a
2:08:54and the very last one we have 108 grams
2:08:56of a g again given mass is 1 0 8
2:08:58molecular mass is 1 0 8 into we will do
2:09:02n a right to get number of atoms
2:09:05so it again comes out to be 1 na so
2:09:09students what we have done here is we
2:09:11need to find the maximum number of atoms
2:09:13and we have found the maximum number of
2:09:14atoms in the case a which is a 2 2na
2:09:18clear now we'll proceed towards our next
2:09:20topic so yes my dear students now we are
2:09:23going to start reactions in solution we
2:09:25are going to see the concentration terms
2:09:27so the very first term is mass percent
2:09:31mass percent or mass percentage or in
2:09:33other terms we defined it as weight by
2:09:36weight percent how do we calculate it
2:09:39according to the name mass percent that
2:09:41means there is a relationship between
2:09:43mass right so the definition for mass
2:09:47percent
2:09:49or i would say
2:09:52weight by weight percent is that
2:09:55mass of
2:09:58a component that we need to calculate
2:10:02is
2:10:03present in
2:10:08total
2:10:10mass of the solution
2:10:17it defines us that how much mass of the
2:10:20component is actually present in the
2:10:22total mass of the solution so if i
2:10:25define its formula what would be the
2:10:27formula of weight by weight percent it
2:10:30is refers to add
2:10:32mass of component
2:10:36divided by
2:10:38mass of solution
2:10:43in 200
2:10:45if i speak in terms of solute and
2:10:48solvent what do you mean by solute and
2:10:50solvent students solute refers to the
2:10:53substance which is present in less
2:10:55amount
2:10:56and
2:10:57the solvent is the one which is present
2:10:59in the more amount solution solvent
2:11:01comprises to form solution so if i speak
2:11:04in that terms i would define mass
2:11:07percent or weight percent as
2:11:10mass of solute
2:11:14divided by mass of solution
2:11:18into 100
2:11:20why do we multiply this into 100 because
2:11:23we are calculating this in the terms of
2:11:25percentage that's why so now let us see
2:11:27one question a solution is prepared by
2:11:30adding 4 grams of a substance a to 16
2:11:34grams of water calculate the mass
2:11:36percentage of solute it is very simplest
2:11:39question students here we need to
2:11:41calculate the mass percentage of solute
2:11:43as i told you the mass of solute is
2:11:46always present in less
2:11:47and the solvent is more like
2:11:51the one that is being dissolved is
2:11:52solute and into which
2:11:54it is being dissolved is solvent
2:11:57for example
2:11:59let me explain you in detail
2:12:01this is my beaker it contains
2:12:03solvent
2:12:06now the thing that i am adding from
2:12:09above
2:12:11is known as solute
2:12:13so when solute is being added into
2:12:15solvent it refers to as solution it
2:12:18consists of solution
2:12:19they are asking us to calculate the mass
2:12:22percentage of the solute that is being
2:12:24added that means it they are asking
2:12:27about the
2:12:29mass percentage of the substance a that
2:12:32is being added into the solution so what
2:12:34is the formula its formula is mass
2:12:37percent
2:12:38or weight by weight percent is equals to
2:12:41mass of component which is the component
2:12:44a 4 grams mass divided by mass of
2:12:46solution now solution comprises of both
2:12:48solute and solvent that means 4 grams of
2:12:51substance a plus 16 grams of water will
2:12:55make a solution
2:12:56into 100
2:12:58so it is 4 by 20 into 100 now four fives
2:13:04is 20
2:13:055
2:13:0620 size 100 hence answer is 20 percent
2:13:10that means the mass of the solute is 20
2:13:13percent it was an easy question it was
2:13:16an easy formula so now students we are
2:13:19considering one more example here see a
2:13:21solution contains 20 grams of glucose in
2:13:24100 grams of water calculate mass
2:13:27percentage of glucose as well as water
2:13:29now see we need to calculate the
2:13:32mass percentages of both the components
2:13:35of glucose as well as the water so the
2:13:36first formula will work here like the
2:13:39component whose mass you have to
2:13:40calculate divided by total mass into 200
2:13:43so it is as simple as that first of all
2:13:46you need to write the mass
2:13:48percentage of glucose to calculate the
2:13:51mass percentage of glucose you need to
2:13:53calculate
2:13:55the mass of
2:13:56glucose upon total mass of the solution
2:14:02total mass
2:14:03of the solution
2:14:05and into 100 this will calculate for
2:14:08glucose now if i ask about mass
2:14:11percentage of water what will be the
2:14:14formula it will be mass of water
2:14:19divided by total mass
2:14:23of solution students into 100 now the
2:14:28component whose whosoever you want to
2:14:30calculate should be kept on the above on
2:14:33the numerator and on the denominator
2:14:35always you have to
2:14:37put the total mass of solution please
2:14:39remember that you have to put up
2:14:40solution not of the particular solvent
2:14:43sometimes students only write what is
2:14:45given in the question they forget that
2:14:47solution is uh consisting of solute as
2:14:50well as solvent right
2:14:52so now if i put into these situations
2:14:55here
2:14:56mass of glucose is given 20 grams and
2:14:58solution mass is going to be the
2:14:59addition of both of them that is 20 plus
2:15:02100 into 100 so when we solve it it is
2:15:0620 by 120 into 100 right
2:15:11ah i'm cutting the zeros then
2:15:142 6 are 12 here i'm getting
2:15:17100 by 6 which is going to be 16.67
2:15:21percent something
2:15:23it is 16.67
2:15:25percent so the percentage mass of
2:15:27glucose is 16.67 percent now i'm going
2:15:31to calculate the mass of water so the
2:15:33mass percentage of water is
2:15:35is mass of water divided by total mass
2:15:38into 100 students so the mass of water
2:15:41what it is 100 grams given in the
2:15:43question
2:15:44right students it is already
2:15:47given to us in question divided by total
2:15:50mass of the solution solution consists
2:15:52of solute as well as solvent so we need
2:15:53to add both both of them that is 20 plus
2:15:57100 into 100 so here what i get 100 upon
2:16:01120 into 100 again one zero is calc
2:16:05canceled with the other zero now if i
2:16:08cut it on the table of two
2:16:11it is 500 by six students so i'm getting
2:16:15it's it 83.33 percent something
2:16:21right so this was all about the mass
2:16:24percentage question now students we are
2:16:26moving towards the next concentration
2:16:28term this that is referred to as mole
2:16:30fraction it is the most simplest
2:16:32concentration term i would say mole
2:16:34fraction what do you mean by mole
2:16:37fraction first of all understand this
2:16:39what is moles moles refers to as given
2:16:42mass
2:16:45upon molecular mass right
2:16:47okay now to define
2:16:51mole fraction in detail and to discuss
2:16:54about moles in detail students first of
2:16:57all
2:16:58you need to understand
2:17:00about
2:17:02mole fraction mole fraction is
2:17:04represented by zeta
2:17:07it is equals to number of moles of
2:17:09substance a divided by number of moles
2:17:12of substance a plus number of moles of
2:17:15substance b that means total substance
2:17:18moles into 100
2:17:21right so students i'm considering that
2:17:24um there are two beakers or it is
2:17:26divided into uh like half of the
2:17:29category so one contains a particles
2:17:32another contains b particles now if i
2:17:34have to calculate how much is a
2:17:36particles present in that beaker i need
2:17:38to calculate number of moles to
2:17:40calculate number of moles for the
2:17:41particular component you have to keep
2:17:43the number of moles of that component on
2:17:45the above position and it should be
2:17:47divided by total number of moles so
2:17:49hence if i have to calculate the mole
2:17:50fraction of a
2:17:52i will keep number of moles of a above
2:17:54divided by total number of moles in 200
2:17:57right now moving towards the mole
2:17:59fraction of
2:18:04b
2:18:04what will be the formula number of moles
2:18:06of b divided by total number of moles
2:18:10into 100 now i have already written this
2:18:13formula also in case in some of the
2:18:16questions what happens is
2:18:17there are
2:18:18they don't give you directly moles they
2:18:20give you given mass as well as molecular
2:18:22mass so first of all you need to
2:18:23calculate number of moles with the given
2:18:25formula given mass upon molecular mass
2:18:27and then you need to apply it into the
2:18:29following formulas that's it students so
2:18:32students now uh let us see an example
2:18:35for this
2:18:36if 30 grams of acetic acid is mixed with
2:18:4046 grams of ethanol then find mole
2:18:43fraction of both the compounds now for a
2:18:46as well as for b that means for acetic
2:18:49acid as well as for ethanol we need to
2:18:51calculate the mole fraction so first of
2:18:53all acidic acid
2:18:57is
2:18:58ch3cooh
2:19:02right and ethanol
2:19:05formula is c2h5oh
2:19:09no need to worry that mam we don't know
2:19:11the formula no need to worry
2:19:14you inorganic chemistry when you come
2:19:16across the chapters you will learn these
2:19:18concepts also what is the formula how is
2:19:21the composition you'll get to know in
2:19:23that terms first of all i'm already
2:19:24telling you because you are new to these
2:19:27terms right
2:19:28so acetic acid is ch3coh while ethanol
2:19:31is c2h5oh now we need to calculate the
2:19:33mole fractions so we need to calculate
2:19:35the molecular mass why i am saying we
2:19:37need to calculate molecular mass because
2:19:39students here number of moles are not
2:19:41given to us directly so first we need to
2:19:43calculate number of moles for the
2:19:44particular components and for that given
2:19:46mass is already given in the question
2:19:48now we will calculate the molecular mass
2:19:51students
2:19:52right
2:19:53to calculate the molecular mass what we
2:19:55are going to do see how many carbons are
2:19:57there two carbons two into carbons mass
2:20:00that is carbon mass is 12 plus how many
2:20:03hydrogen four four into hydrogen mass
2:20:05hydrogen has one right plus um two
2:20:09oxygen atoms into one mass of oxygen
2:20:11atom is 16 so what will be the molecular
2:20:14mass 24 plus 4 plus 32 that is going to
2:20:18be 28 plus 32 that means it is 60 grams
2:20:22this is the molecular mass for acetic
2:20:25acid now let us calculate its number of
2:20:28moles
2:20:29number of moles for
2:20:32acetic acid is
2:20:33given mass upon molecular mass what is
2:20:36the given mass in the following question
2:20:3830 grams divided by molecular mass is
2:20:4160. so i'm getting number of moles as 1
2:20:44by 2 right now moving towards the
2:20:46ethanol portion first of all we need to
2:20:48calculate the molecular mass what will
2:20:50be the molecular mass for this how many
2:20:52carbons 2
2:20:54the mass for one carbon atom is 12 so
2:20:57we'll multiply it with 12 plus how many
2:20:59hydrogen 6 one is the mass of hydrogen
2:21:02plus
2:21:031 into oxygen's mass that is 16. so it
2:21:06is going to be 24 plus 6 plus 16 that is
2:21:09going to be 46 students so this is the
2:21:12molecular mass for ethanol now we will
2:21:15calculate number of moles for ethanol
2:21:17ethanol has molecular formula c2h5oh so
2:21:20i'm writing is like this
2:21:22given mass upon molecular mass gives us
2:21:24number of moles what is the given mass
2:21:26given in the question 46 what is the
2:21:28molecular mass we have calculated 46 so
2:21:31how many moles one moles now to
2:21:33calculate mole fraction we have already
2:21:35calculated number of moles we will apply
2:21:37our basic formula what is the basic
2:21:39formula
2:21:40number of moles of a component divided
2:21:43by total number of moles
2:21:45is equal to number of moles for the a
2:21:48component let us say
2:21:50this is the a component
2:21:52and this one is the b component right so
2:21:55okay first of all number of moles of a
2:21:57component is half
2:21:59divided by total moles that is going to
2:22:01be half plus one
2:22:03so it is going to 1 by 2 divided by
2:22:083 by 2 hence it is coming out to be 1 by
2:22:113 so the mole fraction i would say for
2:22:16ch3coh is going to be 1 by 3 now to
2:22:19calculate for ethanol we have to do same
2:22:22mole fraction is going to be number of
2:22:24moles of b divided by total number of
2:22:26moles what are the number of moles of b
2:22:28they are one divided by what are the
2:22:31total it's 3 by 2 as 1 by 2 plus half 1
2:22:36by 2 plus
2:22:371 is 3 by 2 so i've directly written
2:22:39that 3 by 2 so 2 will go upwards it will
2:22:42be 2 by 3 hence mole fraction for
2:22:45ethanol is 2 by 3 and for acetic acid is
2:22:49one by three so this is the um formula
2:22:52how we can calculate mole fraction right
2:22:54students now we are moving towards our
2:22:57next concentration term that is known as
2:22:59ppm it is very simplest what is ppm it
2:23:01is the concentration of component
2:23:04basically the component whose
2:23:06concentration we need to calculate
2:23:09present in 10 raised to the power six
2:23:11parts of solution first of all write the
2:23:14full form of this
2:23:16parts
2:23:17per
2:23:18million
2:23:20as you can know here it is written parts
2:23:23per million that's why we have taken 10
2:23:25raised to the power 6 here so what is
2:23:27going to be the formula for ppm to
2:23:29calculate ppm the formula is mass of
2:23:32solute
2:23:35divided by
2:23:36mass of solution
2:23:39mass of solution students into 10 raised
2:23:43to the power 6.
2:23:45now let us
2:23:46one second solution will come here
2:23:48kindly correct that um just think that
2:23:51when we were doing mass percentage what
2:23:53we were doing mass of solute divided by
2:23:55mass of solution in 200 but now we are
2:23:58taking mass of solute divided by mass of
2:24:00solution into 10 raised to the power 6
2:24:02what is the basic difference between
2:24:04both of these here we are talking about
2:24:07parts per million that's why we are
2:24:10melting it with 10 raised to the power
2:24:126. this is the basic difference that's
2:24:14all students so this formula is being
2:24:16applied to calculate the percentage ppm
2:24:18that is parts per million percent so now
2:24:21we will move forward towards our next
2:24:23concentration term that is molarity and
2:24:25students um you all will be thinking
2:24:27that ma'am so many formulas how will we
2:24:31we will learn these formulas so trust me
2:24:33these formulas can only be learned by
2:24:36your own practice
2:24:37practice determination and consistency
2:24:41these three things will lead you towards
2:24:44success and if you will follow these
2:24:47three things then you can achieve
2:24:49anything see for the very first day when
2:24:51you start doing these questions and when
2:24:53you
2:24:53listen to our lectures some of the
2:24:55students will grasp it quickly while
2:24:58some of the students will feel like it
2:25:00is taking time for us
2:25:01no need to worry at all just three
2:25:04things you need to do is you need to
2:25:06increase your day to day time schedule
2:25:08if you are working for one hour you are
2:25:10studying for one hour increase it to one
2:25:13and a half hour then increase it to two
2:25:15hours then on the next day increase it
2:25:17to two and a half hour like this you
2:25:19will be able to sit and do your work and
2:25:23do practice and the consistency will
2:25:25increase and maintain so this is the
2:25:27basic thing don't start in a rush that i
2:25:30i have to do complete in a day i have to
2:25:33do complete slabs in a day no no
2:25:35take small small steps and then you will
2:25:38achieve to your success right as you
2:25:41know
2:25:42um like if there are two students you
2:25:45know there is two way to achieve to
2:25:47success one is stairs another one is
2:25:49escalator or i would say lift i would
2:25:51say so um
2:25:53so the stairs won't uh
2:25:57stop you to achieve to success you can
2:25:59go through stairs but it will take time
2:26:01it will take time small small small
2:26:03small little little steps but what if
2:26:05lift uh falls or lift stops in between
2:26:09towards moving from ground towards the
2:26:11height what happens what you will do you
2:26:13can't do anything so always the method
2:26:16which is hard try to do that
2:26:19no no problem that it is taking time
2:26:22there is no problem you have time and
2:26:23you can do so do and you will achieve
2:26:26success right so small small efforts
2:26:28leading towards molarity what is
2:26:29molarity molarity is the simplest term
2:26:33simplest concentration terms which
2:26:35defines us that number of moles of
2:26:37salute present in 100
2:26:40volume of solution 100 liters of
2:26:43solution clear students so students
2:26:45molarity is symbolized by capital m
2:26:49right
2:26:50we symbolize molarity by capital m what
2:26:54is molarity basically molarity is number
2:26:57of moles of solute present in one liter
2:27:02of solution so what will be the formula
2:27:05of molarity
2:27:07molarity is
2:27:09number of
2:27:11moles of solute
2:27:16present in
2:27:18present in one liter of solution that
2:27:20means we have to take volume of solution
2:27:23below
2:27:24and it should be in liters
2:27:27right if you are taking volume of
2:27:30solution in milliliters the formula will
2:27:32be modified as molarity
2:27:35is equals to
2:27:37number of
2:27:40moles of solute
2:27:42divided by
2:27:45volume of solution
2:27:48in milliliter hence it will be divided
2:27:51by thousand if you are taking in
2:27:53milliliter terms one thing more students
2:27:55the twist that comes in the questions is
2:27:57that they don't give you directly number
2:27:59of moles they will give you given mass
2:28:02so you have to replace number of moles
2:28:04by given mass upon molecular mass they
2:28:06can give you direct but in some cases
2:28:08they don't give you number of moles they
2:28:10give you given mass molecular mass you
2:28:12need to calculate and put it into the
2:28:13formula so this was all about molarity
2:28:16students in molarity one of the
2:28:18questions is in front of you find a
2:28:20molarity of 40 grams naoh
2:28:23dissolved in 250 ml solutions so listen
2:28:27what is given mass here 40 grams as i
2:28:29told you they will not give you
2:28:31number of moles directly so given mass
2:28:34is 40 grams
2:28:36okay now
2:28:37naoh what with molecular mass molecule
2:28:41and mass for naoh how many sodium one
2:28:43sodium mass is 23 oxygen mass is 16
2:28:47hydrogen mass is one hence the molecular
2:28:50mass is 40 grams so now we need to
2:28:53calculate molarity what is the formula
2:28:55of molarity molarity is number of moles
2:28:57what is number of moles given mass upon
2:28:59molecular mass given mass is 40
2:29:02molecular mass is 40 divided by volume
2:29:04of solution in milliliters now when we
2:29:07take in milliliters we divide it by
2:29:09thousands so 250 divided by 1000 40 is
2:29:13cut by 40 and
2:29:15then it is again cut hence we get 4
2:29:19molar students we are getting four molar
2:29:22right okay
2:29:24now students
2:29:26what we need to do is
2:29:28we have to do molarity of dilution what
2:29:31do you
2:29:34see
2:29:42three
2:29:43five
2:29:50[Music]
2:29:59now we are going to do molarity of
2:30:02dilution students dilution means to add
2:30:04something right if i am having one
2:30:07beaker what i'm doing i am having two
2:30:10beakers right
2:30:11i have two beakers now
2:30:14in this beaker students
2:30:16this is my volume
2:30:18v1
2:30:20it has
2:30:23volume v2
2:30:24and this has number of moles as n and
2:30:28this has number of moles as n what
2:30:30happens is initially i had this beaker i
2:30:33just added some amount into this and so
2:30:37it converted into volume v2 but number
2:30:41of moles remained same y number of moles
2:30:44remained same because number of moles is
2:30:46depended upon given mass and molecular
2:30:48mass that's why and we are adding
2:30:50something and number of moles you know
2:30:53what is molarity's formula molarity
2:30:55formula is number of moles of solute
2:30:57divided by volume of solution and when
2:31:00we are adding liquid then volume is
2:31:02being increased not number of moles
2:31:03number of moles of solute remains the
2:31:06same because we are taking here solute
2:31:08moles not solution moles if we used to
2:31:11take solution moles then it would had
2:31:13increase but we are thinking of solute
2:31:16so
2:31:16n stays remain in both of the cases
2:31:19while volume changes from v1 towards the
2:31:21v2 now what happens is students when n
2:31:24is same for both of the cases what we
2:31:26say n is equals to mv now what we can
2:31:29say if n is equals to mv
2:31:31then we can write n is equals to m1 v1
2:31:35and n is equals to m2e2 now both of the
2:31:38terms will be equal hence the formula
2:31:40obtained is m1v1 is equals to m2v2
2:31:43this is the formula for
2:31:46molarity
2:31:47in dilution molarity of dilution that
2:31:49means for example you are given a
2:31:51question as you can
2:31:53see on the screen how many volume of 10
2:31:56molar hcl solution
2:31:58should be dilute to water to prepare 2
2:32:01liter of 5 molar hcl now let us see
2:32:04within help of diagrams to make it more
2:32:07understandable i have these two beakers
2:32:09in the very first beaker
2:32:12they are saying how many volume of 10
2:32:14molar hcl i have molarity here
2:32:1810 molar right
2:32:20should be dilute to water
2:32:22should be dilute
2:32:24what should we add into it to prepare
2:32:27what we need to prepare what we are
2:32:29preparing students you need to check
2:32:31this what we are preparing one second
2:32:34yes i am adding something into this and
2:32:37it converts into this
2:32:39and to prepare 2 liter off now it is
2:32:42becoming 2 liter
2:32:44and 5 molar hcl now what i've got i have
2:32:48got this volume v2 i have got this
2:32:51molarity m2 and the start i have
2:32:54molarity m1 while i don't have v1 right
2:32:57so first of all i will calculate v1
2:32:59right see
2:33:01so m1 v1 is equals to m2 v2 m1 i know is
2:33:0610 v1 i don't know is equals to m2 is 5
2:33:09and v2 is 2
2:33:11v1 is equals to 5 into 2 by 10 hence v1
2:33:15is 1 liters that means that means the
2:33:19first beaker had one liter volume but
2:33:22the question is incomplete now they are
2:33:25asking us how many volume of 10 molar
2:33:27hcl should we dilute it should be dilute
2:33:30how much should be added so that it
2:33:33becomes 2 liter what should we add in 1
2:33:35liter so that it becomes 2 liter
2:33:37obviously 1 liter will be added into 1
2:33:39so how many volume will be added 1 liter
2:33:42volume will be added but if it asks you
2:33:45in question what is the volume of beaker
2:33:49first or before dilution then you would
2:33:51say 1 liter and when it asks you how
2:33:55much dilution is required then you will
2:33:57subtract your v2
2:33:59right v1 from v2 and you will get your
2:34:01answer
2:34:02right students okay now comes molarity
2:34:05of mixture
2:34:07molarity of mixture refers to as when
2:34:08one beaker is mixed into another beaker
2:34:11and we are getting a mixture
2:34:13for example
2:34:14i have this speaker
2:34:16i have this beaker
2:34:19then students i have the speaker right
2:34:22see
2:34:22i am adding
2:34:24these two and i'm getting this product
2:34:26right this is my beaker a this is my
2:34:28beaker b and this is my solution
2:34:32which i am getting after the addition of
2:34:34both of them here in a molarity was m1
2:34:38volume was v1 here molarity was m2 and
2:34:42volume was v2
2:34:44here
2:34:45molarity will be m refers to as the
2:34:49total molarity of the solution while
2:34:51volume will be
2:34:53v 1 plus v 2 right so what will be the
2:34:56formula to calculate the solutions
2:35:00molarity the complete uh formula
2:35:04will be m1 v1 plus m2 v2 now to
2:35:09calculate molarity of the solution
2:35:11molarity of the solution that is capital
2:35:13m is equals to m1 v1 plus m2 v2 divided
2:35:17by total volume that is v1 plus v2
2:35:22hence this is the formula that is used
2:35:24to calculate the molarity of mixtures
2:35:27clear students very good
2:35:29now students we are moving towards the
2:35:32relationship between molarity and
2:35:34molality
2:35:36what is the relationship between
2:35:37molality and molarity before that let us
2:35:40understand the basic difference between
2:35:43molality molality is that concentration
2:35:46term in which we don't use volume
2:35:49students in molarity what we used to do
2:35:52we used to define
2:35:54molarity
2:35:55as number of moles of solute
2:35:59divided by volume of solution in liters
2:36:03this was molarity but when i talk about
2:36:05molality what do you mean by molality
2:36:08molality is number of moles
2:36:13of solute
2:36:16present in
2:36:20number of moles of solute present in
2:36:23mass of solvent in kg
2:36:26now
2:36:27one more difference which you will
2:36:29observe is in molarity we were taking
2:36:31volume of solution but in molality we
2:36:35are taking mass of solvent and in kg so
2:36:37formula will be molality is represented
2:36:40by small m it is
2:36:42number of moles of
2:36:45solute
2:36:47divided by mass of solvent
2:36:54in kilograms
2:36:56so this is the basic formula that you
2:36:58need to remember now the relationship
2:37:00between molality and molarity what is
2:37:02that this is the relationship that you
2:37:04need to know molality is equals to
2:37:07molarity divided by d refers to as
2:37:09density minus molarity into molecular
2:37:12mass divided by 1000 this is the direct
2:37:15formula that you can apply in the
2:37:16questions when there is a relationship
2:37:18when you are being asked about morality
2:37:20and you have been given molarity so you
2:37:22can directly equate and get your answers
2:37:25clear students so this was the most
2:37:27interesting concentration terms as there
2:37:29is a relationship try to solve as many
2:37:32questions as you can so that the
2:37:33relationship formula stays in your mind
2:37:35right okay now students comes the next
2:37:39concentration term that is referred to
2:37:40as normality
2:37:43to understand normality first we need to
2:37:45understand what is equivalent mass
2:37:48what is equivalent mass students
2:37:50equivalent mass basically is molecular
2:37:55molecular weight by valency factor
2:37:59equivalent
2:38:04equivalent mass is equals to
2:38:07molecular mass or molecular weight
2:38:11basically divided by valency
2:38:14factor
2:38:16valency factor or x you can represent it
2:38:20so students now
2:38:23what is valency factor
2:38:29see for acids
2:38:32if i talk about acids bases and salt
2:38:36first we need to calculate valency
2:38:38factor for acids we always go for
2:38:41basicity for base we always go for
2:38:43acidity man what are you seeing what are
2:38:45acidity what is basicity how do we
2:38:48identify it so students first of all you
2:38:50need to understand the basic formula
2:38:52that is going to help you to calculate
2:38:54normality first you will calculate
2:38:56equivalent mass how to calculate
2:38:58equivalent mass molecular weight by
2:39:00valencia factor then students valencia
2:39:03factor for acids the formula will be
2:39:05changed formula for acids will be
2:39:08equivalent
2:39:12mass
2:39:14is equals to molecular weight
2:39:18by
2:39:18basicity what is basicity basicity is
2:39:22the number of h h-positive ions that are
2:39:24liberated for example students now let
2:39:27us take an example to understand the
2:39:28situation um
2:39:30you can take h2so4 i guess and other
2:39:34than this
2:39:35you can take any of the acids right and
2:39:38for base students
2:39:40first let us write all the um formulas
2:39:42then i'm going to take some of the
2:39:44examples right
2:39:45for acids for base for base the formula
2:39:49is going to be
2:39:50equivalent mass
2:39:56is equals to
2:39:58molecular weight
2:40:01by
2:40:02acidity
2:40:05students acidity refers to as how much o
2:40:08h negative ions are being liberated for
2:40:10example in na which how many o h ions
2:40:13are liberated one right so basis its
2:40:16acidity will be one you will write one
2:40:19here and for the following question you
2:40:20will write the molecular weight and you
2:40:22will calculate the equivalent mass but
2:40:24if i talk about salt if i talk about
2:40:27salt students in the case of salt what
2:40:29happens is
2:40:31equivalent
2:40:34mass
2:40:36is equal to molecular weight
2:40:40divided by what now you have to see the
2:40:43charge you have to see either the charge
2:40:45of cation or the charge of n9 any one of
2:40:48these right
2:40:55cation or a9 for example mgso4
2:40:59in mgso4 divide it it will split into mg
2:41:03plus 2
2:41:05and so4 2 negative so you either take
2:41:08the charge of the positive or the
2:41:10negative it's 2 so its valency factor
2:41:12that the charge of cation 9 will be 2.
2:41:15don't put the charges positive negative
2:41:16that won't be put it only you have to
2:41:18put the number that is true so valencia
2:41:21factor in this case will be 2. now for
2:41:23basicity as i've given the example of
2:41:24nuh how many ohms are liberated only 1
2:41:27then its acidity will be one similarly
2:41:30for acids you can go for like basicity
2:41:32basically refers to as how many h
2:41:34positive ions are liberated that will
2:41:36tell you it's about basicity so this was
2:41:39all about valency factor which will lead
2:41:41you towards the equivalent is and when
2:41:44you have already calculated equivalent
2:41:46weight we will move towards the formula
2:41:49of normality students now students
2:41:52number of gram equivalent we have to
2:41:54calculate number of gram equivalent what
2:41:56is this
2:42:02number of
2:42:05gram equivalent we have calculated
2:42:08equivalent mass equivalent weight so
2:42:10first of all we will write given mass
2:42:14divided by the one that we calculated
2:42:16before that is
2:42:18equivalent weight
2:42:21and you know what is the formula of
2:42:22equivalent weight molecular weight
2:42:25divided by valency factor so first you
2:42:27will calculate this then you will
2:42:29calculate number of gram equivalent and
2:42:31now we will calculate our normality
2:42:34students what is the formula for
2:42:36normality
2:42:41number of gram equivalent
2:42:47number of gram equivalent this one
2:42:51this one number of gram equivalent
2:42:53divided by volume of solution in liters
2:43:03in liters and if they ask in milliliters
2:43:06then you need to divide it by thousand
2:43:08so students it was about normality and
2:43:11when so ever they ask you relationship
2:43:13between normality and molarity what is
2:43:15the basic formula normality is equals to
2:43:17molarity into its atomicity that means
2:43:20valency factor
2:43:23which we calculate for acid bases and
2:43:26salt
2:43:27so this is the relationship between
2:43:29normality and molarity clear we have
2:43:32done two relationships between molality
2:43:34and molarity now we have done between
2:43:36normality and molarity clear students
2:43:39very good
2:43:40now students the next thing that comes
2:43:42is the relationship between molarity and
2:43:44normality i've already explained you in
2:43:45the previous slide i'll proceed towards
2:43:48the next topic that is percentage
2:43:49composition now percentage composition
2:43:52is required for empirical and molecular
2:43:55formula first of all we need to
2:43:57calculate we have been given empirical
2:43:59formula then we will calculate its
2:44:01molecular formula so before
2:44:02understanding into these terms the basic
2:44:05terms that will help us to calculate
2:44:07that is the percentage composition now
2:44:10let us start this so yes my dear
2:44:12students now we are doing percentage
2:44:14composition basically percentage
2:44:16composition is going to help us in
2:44:18empirical and molecular formula what is
2:44:21percentage composition for any component
2:44:23we defined it as mass of that component
2:44:26divided by total mass of the solution
2:44:29into 100 so the percentage composition
2:44:36of any component
2:44:40is equal to the mass of component
2:44:47whosoever we need to find out the
2:44:49percentage that will be written on the
2:44:51above divided by mass of the compound
2:44:55full compound into hundred
2:44:58now let us take an example for the
2:45:01students we have the example calculate
2:45:03percentage composition of carbon and ch4
2:45:06so it is very simple first of all
2:45:08percentage of carbon in ch4 will be mass
2:45:14of carbon
2:45:18divided by total mass
2:45:21total mass of compound what is the
2:45:23compound here ch4
2:45:25into 100 so now students what is the
2:45:28mass of carbon it's 12 what is the total
2:45:31mass ch4 molecular mass it is 12 plus 4
2:45:34right
2:45:35into 100 hence it will come out to be 12
2:45:39divided by 16 into 100 this is the
2:45:42percentage composition for carbon in ch4
2:45:46students we can further divide it like
2:45:48if we go on two table then it is divided
2:45:52further
2:46:00it is 75
2:46:03that means carbon is present 75 percent
2:46:05in ch4 and once they ask you about one
2:46:08component they can even ask you about
2:46:10another one for example hydrogen so
2:46:12hydrogen you can calculate in the
2:46:14similar way like you did in the carbon
2:46:16we first you will keep the mass of
2:46:17component of hydrogen divided by total
2:46:19mass in 200 now this is going to help us
2:46:22in empirical and molecular formula what
2:46:24do you mean by empirical and molecular
2:46:26formula see whenever a compound we
2:46:29observe has two formulas the empirical
2:46:32formula represents the smallest whole
2:46:35number ratio
2:46:36smallest whole number issue this is the
2:46:38most important line that you need to
2:46:40remember
2:46:41whenever you see a compound a compound
2:46:44can be represented in two categories
2:46:46empirical as well as molecular formula
2:46:48so in empirical we need to see the
2:46:50smallest whole number ratio for example
2:46:52this is our compound what will be the
2:46:54smallest whole number ratio that the
2:46:56number through which they can be
2:46:57divisible like 6 12 and 6 they can be
2:47:00divisible on six table so if i divide
2:47:03this then it will be c 1
2:47:05h h2o1 that means ch2 oh
2:47:09ch2o is the simplest whole number ratio
2:47:12that means
2:47:14for this compound this is its empirical
2:47:16formula clear now what do you mean by
2:47:19molecular formula molecular formula
2:47:21represents the actual number of
2:47:23individual atom
2:47:25in any molecule for example in c6h12o6
2:47:28what is the actual number of carbon 6
2:47:31what is the actual number of hydrogen 12
2:47:33what is the actual number of oxygen six
2:47:35so molecular formula represents actual
2:47:37number of the each individual atom while
2:47:40its simplest whole number ratio
2:47:41represents its empirical formula now
2:47:44students what is the relationship
2:47:45between empirical formula and molecular
2:47:47formula so i would say n is equals to
2:47:51molecular formula
2:47:57divided by empirical formula
2:48:05so this is one of the uh relationship
2:48:07what we can say it molecular formula we
2:48:09can take molecular weight instead of
2:48:11this and in empirical formula we can
2:48:13take empirical weight instead of this
2:48:15because we need to calculate weight here
2:48:18for both of these right okay now i'm
2:48:21taking an example for you people where
2:48:23we are going to use this formula the
2:48:25question is a compound is molecular
2:48:28weight 90 right its empirical formula is
2:48:31cho2 find the molecular formula now
2:48:34students we have been given empirical
2:48:37formula
2:48:38we have been given empirical
2:48:41formula empirical formula is cho2 what
2:48:45will its empirical weight
2:48:48empirical weight
2:48:50how many carbon one so one into carbons
2:48:53mass plus one into hydrogen mass plus
2:48:56two into oxygen mass carbon mass is 12
2:48:59hydrogen mass is 1 and oxygen mass is 16
2:49:03so when i calculate it it's 12 plus 1
2:49:06plus 32 that means 13 plus 32 it means
2:49:1245 45 is the empirical weight right we
2:49:16have calculated empirical weight from
2:49:19empirical formula but we need to uh
2:49:21calculate molecular formula we have
2:49:23molecular weight with us we have
2:49:26molecular weight
2:49:29which is 90 with us now i have given you
2:49:31a formula which is n is equals to
2:49:34molecular weight 90 divided by empirical
2:49:37formula or empirical weight that is 45
2:49:39what comes out to be n and comes out to
2:49:42be two so what we will do now we have
2:49:45our empirical formula which is cho2
2:49:49which is cho2 so i will multiply
2:49:53i am going to multiply this with 2 so
2:49:56that i can get my
2:49:58empirical sorry so that i can get my
2:50:02molecular formula
2:50:07because i told you in the previous case
2:50:09same this question
2:50:10there are two kinds of formulas one is
2:50:13molecular formula divided by empirical
2:50:14formula the other one is molecular
2:50:16weight divided by empirical weight we
2:50:17have used both of the formulas first we
2:50:20use the weight one then we use the
2:50:22molecular formula one so what
2:50:24will it comes out to be
2:50:26it comes out to be c2 h2o4
2:50:31so this is going to be the molecular
2:50:34formula like this you need to put it
2:50:36into the down at the base not on the top
2:50:38you have to write it c2 h2o4 because
2:50:42here also in the previous case i told
2:50:43you now what is the number that divides
2:50:45it into simplest whole number ratio
2:50:47right so we are dividing it that is n
2:50:50right so here this is the complete
2:50:53number of atoms it represents the
2:50:55complete number of atoms while it
2:50:56represents the simplest whole number
2:50:58ratio so this is empirical formula this
2:51:00is our molecular formula so you can
2:51:02calculate these kinds of questions in
2:51:04such a way now students proceeding
2:51:06towards the next kind of question what
2:51:08you need to do is most of the times
2:51:09these questions are observed a compound
2:51:12contains 38.8 carbon 16 hydrogen and
2:51:1742.2 45.2 percent nitrogen the formula
2:51:22of the compound would be first of all
2:51:24you need to write element
2:51:26first of all you will write about
2:51:28element how many elements are present
2:51:30this carbon hydrogen nitrogen the next
2:51:34thing students what you need to write
2:51:36down is
2:51:38the next thing you will write the
2:51:40percentage composition or you can write
2:51:41atomic mass first so i am going to write
2:51:44atomic mass first of these compounds
2:51:47what is the atomic mass of carbon 12
2:51:50what is the atomic mass of hydrogen 1
2:51:52what is the atomic mass of nitrogen it's
2:51:5414 right now the next thing that we are
2:51:56going to write is the percentage
2:51:58composition that is given to you in the
2:52:00question
2:52:02percentage
2:52:05composition now what is the percentage
2:52:07composition for the case of carbon it is
2:52:09given as
2:52:1038.8 percent what is the percentage
2:52:13composition for hydrogen it is given to
2:52:15us 16 percent what is the percentage
2:52:17composition of nitrogen it is given to
2:52:19us
2:52:2045.2 percent students now the next step
2:52:24is going to
2:52:25going to be that we need to find the
2:52:26stoichiometry that is we need to divide
2:52:29percentage composition by atomic mass so
2:52:31what will come let us see
2:52:35we will divide percentage
2:52:40composition by
2:52:42atomic mass
2:52:46so
2:52:47in the case of carbon it will be
2:52:5038.8 divided by 12 for the case of
2:52:53hydrogen it will be 16 by 1
2:52:56now for next case it will be 45.2
2:53:00divided by
2:53:0140. now students when we are going to
2:53:03solve it out it comes out to be
2:53:063.23 something it comes out to be 16
2:53:10students then um here it comes out to be
2:53:13almost 3.232 to something only
2:53:19right so we have just calculated these
2:53:22now we have to put into the simplest
2:53:24form students how to put them into
2:53:26simplest form let us divide it in such a
2:53:28way that we get the answers like the
2:53:31first one we have done 3.23 then we have
2:53:34done 16 then is 3.23 what is the number
2:53:37that divides all of these the number
2:53:39that divides all of these here it would
2:53:41come out to be
2:53:44it would come out to be 1 5 and 1
2:53:50right okay so if you multiply 5 with
2:53:53this you will approximately get
2:53:55let me tell you the complete solution so
2:53:57that you won't be able to find any
2:54:00problem in this
2:54:02see it is
2:54:0316.15 so it is approximately 16 that way
2:54:06we always take at the approximate
2:54:08situation because it's the percentage
2:54:09composition so this is our simplest
2:54:12ratio
2:54:17this refers to as the simplest ratio
2:54:19what does we do when we get this
2:54:20situation we divide the smallest number
2:54:22with everyone we are dividing 3.2 with
2:54:243.23 we get 1 we divided 16 by 3.23 we
2:54:28get 5 we are dividing 3.23 divided by
2:54:31again
2:54:323.23 so we are getting one the smallest
2:54:34number is
2:54:35is being divided by all of them so this
2:54:37is the simplest thing that we observe
2:54:39here now what we have to do carbon base
2:54:421
2:54:44hydrogen base 5 nitrogen base one now
2:54:48let us see the examples you know the
2:54:50options here in all of these options
2:54:52they have opened it up so we also need
2:54:54to open it up how many carbon one carbon
2:54:57in this case one carbon is there in this
2:54:59case two carbons are present so b option
2:55:01is incorrect in this also it is
2:55:03incorrect in this case one carbon is
2:55:05there next one nitrogen and five
2:55:07hydrogen now
2:55:08one nitrogen and
2:55:10five hydrogen it is possible okay here
2:55:13it is two to four hydrogen and 2
2:55:16nitrogen it is not possible hence our a
2:55:18option that is
2:55:21ch3
2:55:22nh2 is a totally correct students so
2:55:26these kinds of questions can arise first
2:55:28of all you need to write the atomic mass
2:55:30then you need to write the percentage
2:55:32composition which is given in the
2:55:33question then you need to divide them
2:55:35and after that you need to select the
2:55:37smallest one and divide them so that you
2:55:38can get the simplest number now the
2:55:40simplest number is put into the base and
2:55:43the formula is obtained so students now
2:55:46we are moving towards the stoichiometry
2:55:48what do you mean by the stoichiometry
2:55:50basically stoichiometry deals with the
2:55:53calculation of masses of reactants as
2:55:56well as products in the chemical
2:55:57reaction
2:55:58whosoever mass we want to calculate
2:56:00whosoever volume we want to calculate
2:56:02with the help of stoichiometry we can do
2:56:04this they establishes basically three
2:56:06kinds of relationships the first one is
2:56:08mass mass relationship for example when
2:56:10two reactants react together and they
2:56:12form product so in the question what
2:56:14they'll do they give you the mass of
2:56:16reactants and they will ask you the mass
2:56:18of products so that comes under the
2:56:19category of mass mass relationship when
2:56:22they give you mass volume relationship
2:56:24what do you mean by that one of the
2:56:25question one of the reactant will be
2:56:27given mass one of the reactant will be
2:56:30given as volume and they will ask you
2:56:32about the mass of the product so these
2:56:34kinds of relationships are mass and
2:56:36volume relationship the last kind is
2:56:39volume volume relationship when the
2:56:40reactants volumes are given and the
2:56:42product volume is being asked so
2:56:44students one of the stoichiometric
2:56:46questions in front of you how to do i
2:56:48will tell you
2:56:49if 16 grams of methane is burnt find the
2:56:52mass of co2 produce so see
2:56:55methane is burnt when it is burned right
2:56:58so when methane is burnt
2:57:01what happens
2:57:02what happens is co2 here is produced
2:57:04right so what what we observe here is
2:57:06what we need to find out the mass of cu
2:57:08to produce here we need to find out the
2:57:10situation and also water will be
2:57:12produced here right now the very first
2:57:15thing the very first step that you need
2:57:17to do is balance the equation
2:57:22how to balance the equation write a
2:57:24reactant decide right product this side
2:57:27how many atoms are present at the
2:57:29reactant first of all we take off carbon
2:57:31atom carbon and the reactant side is one
2:57:34in the product side it is also one okay
2:57:36the next one is hydrogen in the reactant
2:57:38side it is four in the product site it
2:57:40is two next is oxygen in the reactant
2:57:43side it is two while in the product side
2:57:46it is three so we need to balance it out
2:57:47that means we need to equalize it in
2:57:49both the ways now students uh first of
2:57:52all we can multiply this uh product side
2:57:54hydrogen into two so that it is equal to
2:57:56four so what i'm going to do i'm writing
2:57:582 here so that it is multiplied now
2:58:01after putting 2 here
2:58:03hydrogen is 4 but oxygen is also 2 that
2:58:05means on the product side we have 2 plus
2:58:072 4 oxygen we have now 4 oxygen with us
2:58:11right now students but on the reactant
2:58:14side 2 oxygen are present that means we
2:58:16need to multiply here with it with two
2:58:19so here it is four now check all the
2:58:22elements that they are equal in both the
2:58:24size in reactant or product then yes it
2:58:26is equalized why did i this this because
2:58:29stoichiometry is the most important role
2:58:31we need to understand to calculate mass
2:58:33mass relationship mass volume
2:58:35relationship volume volume relationship
2:58:36right now
2:58:38students look here here it is one when
2:58:40nothing is given we consider it as one
2:58:42here it is one it means one mole
2:58:46it means two mole
2:58:48it means one mole
2:58:50it means two mole
2:58:52now what do you mean by this these
2:58:55stoichiometry are the relationships that
2:58:57are being established students these
2:58:59stoichiometric basically are the
2:59:01relationships kindly correct this here
2:59:03now hydrogen is also 4
2:59:05right these are equalized here these are
2:59:07also 4 even oxygens are also 4 on the
2:59:10both the sides hence we have done this
2:59:12right 2 plus 2 4 2 plus 2 4 now how to
2:59:15calculate the mass of co2 produced say
2:59:17students what happens is they have given
2:59:20us the mass of methane ch4 and we need
2:59:22to calculate the mass of this so i would
2:59:25say one mole gives one mole of co2 i can
2:59:28say that one ch4 one mole of ch4
2:59:36one mole of ch4 gives
2:59:39one mole of
2:59:40co2 right
2:59:43so
2:59:44their moles are equal and what are moles
2:59:46given mass by molecular mass
2:59:50here it is equal to given mass per
2:59:52molecular mass of c2 what is the given
2:59:54mass here of methane 16. what is the
2:59:56molecular mass of methane ch4 it is 16
2:59:58again it is equals to given mass that we
3:00:01need to calculate divided by molecular
3:00:02mass of what co2 its molecular mass is
3:00:0544 right so it will solve it and cross
3:00:08multiply we get the given mass as 44
3:00:11grams hence we can calculate but most of
3:00:13the students can ask why not we are
3:00:15considering this o2 2 mole gives one
3:00:18mole because they have only given us one
3:00:20situation when two situations are given
3:00:23then we consider about limiting reagent
3:00:25when two of the reactants information is
3:00:27being provided then we first find out
3:00:29what is the limiting reagent and then we
3:00:31do then if only one equation is given to
3:00:35us then we directly do according to the
3:00:36stoichiometry and we apply that if moles
3:00:39are equal then their formulas will be
3:00:41equal and the answer will be calculated
3:00:43so now students we have our topic
3:00:46limiting reagent what is limiting
3:00:49reagent
3:00:50one day me and my brother were sitting
3:00:52and we were very hungry so it was very
3:00:55late night time and my parents were
3:00:57sleeping so i thought uh to eat
3:01:00something and i went to the kitchen i
3:01:01took my brother with me i said what do
3:01:03you want to eat he said me let's have
3:01:05some burger and all you know so but but
3:01:08in the kitchen i observed to make burger
3:01:10we need some buns and we we need some
3:01:12buns or bread i would say and along with
3:01:14that we need some salad to put inside it
3:01:17we need some cheese to put inside it so
3:01:20what we did was we need some buns and
3:01:23including other ingredients we need
3:01:25these ingredients too so
3:01:27to make burger we had only and only two
3:01:31burger two burger breads right and we
3:01:34had ingredient in large quantity but we
3:01:37were like we were only two of the
3:01:39persons but but we were so hungry that
3:01:42we could eat four or five burgers you
3:01:44know so my brother said that i am very
3:01:46hungry i want to eat more so what did
3:01:49this situation explain us that what is
3:01:53limiting reagent in this case limiting
3:01:55reagent is the one which is present in
3:01:57less quantity and excess reagent is the
3:02:00one which is present in more
3:02:01quantitative quantity
3:02:02so according to this i would say that
3:02:04limiting reagent is the reactant which
3:02:06is completely consumed in reaction how i
3:02:08can explain you in the case of kitchen
3:02:11see if there are two breads then only
3:02:14two burgers can be obtained because
3:02:16bread is responsible because it is in
3:02:18limited quantity it will make only two
3:02:21if breads were more and ingredients were
3:02:23less then we could have done uh
3:02:25something more and made something uh
3:02:27different uh you know
3:02:29food ingredients with us but the bread
3:02:32was only two then the burger produced
3:02:34will be only two because it was the
3:02:36limiting region that means limiting
3:02:38reagent is that reagent which decides
3:02:41that how much the product will be formed
3:02:45okay now
3:02:46see
3:02:47it is the reactant which is completely
3:02:49consumed in a reaction let us take an
3:02:50example for example i have 3a plus 4b
3:02:54gives 5c plus 60
3:02:57okay and i am saying there are number of
3:03:00moles of a is
3:03:0310 while
3:03:05number of moles of b is 20. see in the
3:03:09previous case only one situation was
3:03:11given that was the 16 grams of ch4
3:03:15and it was not given that what is the
3:03:17amount of o2 in it now in this case both
3:03:20of the reactant number of moles are
3:03:21given how do we identify that with the
3:03:24help of which we can calculate c and d
3:03:26amount now we need to find out from both
3:03:29of these reactant which is a limiting
3:03:31reagent and which is excess reagent the
3:03:34the one which is limiting reagent will
3:03:36help us to calculate the amount of
3:03:37substances of the product right so see
3:03:40what happens is this is the balanced
3:03:42equation which is given to us now what
3:03:44we need to do is we see that
3:03:473 a and here we see that 4b we have
3:03:50reactant as 3a and we have another
3:03:54reactant as 4b what is this 3 4 5 6
3:03:57these are basically students these are
3:04:00basically the stoichiometric coefficient
3:04:02what we do is we calculate here number
3:04:05of moles
3:04:08by stoichiometry
3:04:14and whose value is less is the limiting
3:04:17reagent now for 3a number of moles of a
3:04:20is 10 divided by stoichiometry is the
3:04:23number in front of the reactant it is
3:04:25three for b the number of moles is 20
3:04:28and the stoichiometric given here is
3:04:30four it is equal to three point three
3:04:32something and it is equal to five so
3:04:35which amount is less obviously this
3:04:36amount is less this amount is less that
3:04:39means it is limiting reagent
3:04:42it is our limiting reagent which will
3:04:44help us to calculate the values of the
3:04:47product now i ask you what is the amount
3:04:49of substance what are the number of
3:04:50moles obtained of the product c how can
3:04:53you calculate will you take will you
3:04:55consider b now no because b is present
3:04:58in excess and we will find question only
3:05:00through the limiting reagent when two of
3:05:02the substances amounts are given to us
3:05:04first we will find limiting region how
3:05:05to find divide number of moles by
3:05:07stoichiometry whose value is less that
3:05:09is your limiting region now we have
3:05:11found already now as you know that this
3:05:14is 3 moles
3:05:16this is 4 moles
3:05:18this is 5 moles
3:05:21this is 6 moles as we have done in
3:05:23stoichiometry now this will help us to
3:05:25find these right so i would say that 3
3:05:27moles of a
3:05:30gives
3:05:31five moles of
3:05:33c right
3:05:34one mole of a
3:05:37will give five by three moles of c
3:05:41right so how many moles of a are given
3:05:43ten moles right so 10 moles of a
3:05:48will give 5 by 3 into 10
3:05:52moles of c right so by these we can find
3:05:56the number of moles it comes out to be
3:05:5816.66
3:06:00value so with the help of limiting
3:06:02region we can calculate any amount of
3:06:04the substance of the products clear
3:06:06students now proceeding towards one of
3:06:09the questions let us see calculate the
3:06:11limiting region and the mole of product
3:06:13formed in the given reaction we have h2
3:06:16plus
3:06:17half o2
3:06:19gives water the first thing that we do
3:06:21is balance the equation yet it is
3:06:23already balanced here in the reactant
3:06:24side there are 2h here it is 2h here
3:06:28o2 is you know 2 into 1 by 2 is 1 and
3:06:32here it is also 1 it is already been
3:06:34balanced what we are given is it is
3:06:36taken 5 moles and it is taken as 2 mole
3:06:39students so
3:06:41what is the basic thing that we need to
3:06:42do is we need to divide number of moles
3:06:44by stoichiometry number of moles by
3:06:47stoichiometric coefficient number of
3:06:49moles are five in this case in
3:06:50stoichiometry when nothing is given it
3:06:52is considered one that means it is five
3:06:55now here what is number of moles 2 what
3:06:57is stoichiometry half
3:07:00so it will go above and it will become
3:07:024. now from both of these which is
3:07:04present in less amount obviously this is
3:07:05present in less amount that means it is
3:07:07referred to as limiting
3:07:10reagent
3:07:12now that means with the help of this we
3:07:14can calculate the mole of product formed
3:07:16in the given reaction now if we see
3:07:18these are
3:07:20these are
3:07:21already given to us as half moles and
3:07:24here it is one mole the stoichiometry so
3:07:27i would say half moles of o2
3:07:30half moles of o2 gives
3:07:33one mole of water
3:07:36so one mole of o2
3:07:40will give
3:07:42half mole of water
3:07:46right
3:07:46and in other terms we need to calculate
3:07:48for two moles so two moles of photo will
3:07:52give
3:07:53half into two moles of water
3:07:56that means how many moles of water are
3:07:58obtained
3:07:59one mole of water is obtained
3:08:04so by these you can calculate any of the
3:08:07questions clear students so yes my dear
3:08:10students this chapter is finished here
3:08:13hopefully you liked the video you liked
3:08:15the chapter now the basic thing that you
3:08:17need to do is practice practice a lot of
3:08:20questions on the topic wise so that you
3:08:23can be perfect in every topic in every
3:08:25aspect of some basic concepts of
3:08:27chemistry i hope so that you like this
3:08:30chapter
3:08:31keep learning thank you so much have a
3:08:33good day

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