L17 : Molality, Molarity - Basic Concepts of Chemistry, Class 11 Video Lecture

hello friends this video on basic
concepts of chemistry part 17 is
Bloodshy by exam for calm no more fear
from exam if for watching this video
please make sure that you have watched
part 1 part 16 reactions in solutions so
if you see you major in tu reactions
which we do in laboratory is carried out
in solution most of the reaction happens
in college in exit and that's why we
will look into some of the concepts for
reactions in solution so reactions in of
a solution or the amount of substance
present in its given volume can we
express in a 4 different ways that is
mass percentage or weight percentage
then have mole fraction then I have
molarity and then I have well a little
so there are 4 ways to represent the
concentration of substance in the
solution so we will go through all this
what is mass but sin or weight person
so if you see mass percent is a one way
of representing concentration of element
in a compound or a mixture and it is
caroled as a mass with the component
divided by total mass of nature for
example if you talk about water molecule
h2o so this guy has two holes of
hydrogen and one mole of oxygen or you
can say that 2 mole of hydrogen anything
but 2 into 1 grams of hydrogen and one
mole of oxygen is nothing but 18 grams
of oxygen so it's 16 grams of oxygen
right because oxygen 16 so you see I'll
have to find mass percentage of hydrogen
that will become 2 by 16 plus 2 18 in
200 anywhere to find mass percentage of
oxygen it will be 16 by total is 80
that's how we find that will be in
percentage so that is nothing but is one
way of representing concentration of
element in a compound or a mixture I
have a compound water I have to find a
percentage of hydrogen oxygen I can find
using this way this is mass percentage
we condemn it examples
of solutions prepared by adding two gram
of substance a - 18 gram of water right
find the mass percent of solid so you
see in this case I am talking about the
percentage of substance in word rates
it's a mixture it is not our components
mixture here so I have two grams of
substance and 18 grams of water eight
total I have how much 20 grams of
mixture is a mixture so I say percentage
of substance there is nothing but 2 by
mixture is 20 in 200 that is 10%
I am talking about so this is my
percentage of substance that is one way
of representing the percentage of
substance in a mixture so in this
mixture a program substance eating ramps
of water total mixture is 20 grams and I
have to find the percentage on my image
a substance and that I found is
two-person this is one way of
representing mass or percentage of a
substance in a solution or a mention the
other thing is a mole fraction mole
fraction is worth the ratio of number of
moles of a particular component the
total number of moles of the solution so
for example I have let's suppose 3 mole
of X and I have that suppose 7 hole of Y
right
I have these two you know in a mixture
or a solution I have a 3 mole of this
and 7 will of Y in a solution now if I
say mole fraction of X that will become
3 by 7 plus 3 is 10 in 200 percent is
that is if you see 30 percent and I will
see mole fraction of Y it will be 7 by
and E if you want to go write in
percentage also or you can write
normally 0.3 here you can write point 7
also
your wish if you want to write terms of
percentages you have to multiply 100
become 30% or you can write in fraction
also but since the faction will will
write in terms of fraction so it is a
mole of X by mole of the whole solution
so this is a mole of X is 3 mole of
oscillation is 10 3 by 10 in fraction in
this case mole of Y is 7 bowl of total
solution is tense in embedded in spine
step now if I write you write the mole
of particular X by Y 2 mole of total
solution right that is mole fraction of
X generals form lines mole fraction of X
is nothing but mole of x by mole off
whole solution correct then we have
something called molarity it is the most
widely used so please please a little
attention here it is noted by capital M
please it is known as capital M is now
small m and it is defined as number of
moles of solute in 1 liter of solution
please know it is number of moles of
solute in 1 liter of solution please
it's again repeat this definition there
is nothing but number of moles of solute
of solute in 1 liter of solution take
one example on this for RAM sugar this
formula lizard in third 350 milliliter
of water
what is the molarity of a solution so
first thing is we have told that moles
of solute per liter right this is this
is capital of molarity so to find first
this volume moles all right so this guy
formula is c12 h22 o11 so let us find
the molar mass of this molar mass of c12
h22 o11 is nothing but C 2 L is 12 into
12 s 2 D 2 is 22 11 into 1 or will ever
this level into 6 that comes out to be 3
42 grams per
more this is the molar mass correct that
means three 42 grams implies 1 mole of
sucrose right but I have four Rams so
four grams implies 4 by 342 mole or
sucrose that is point zero 1 1 7 1 so I
know the mole I was given that sugarless
program I know for gram sugar is this
meaning balls correct so I have this M
polarity is nothing but mole of solute
that is suppose in this case by volume
of solution there going any point zero
zero one seven by this has to be liter
350 ml is nothing but point three five
zero later correct and this comes out to
be point zero three mole per liter mole
per liter is a unit of molarity and that
is my answer he has simple so you have
to find a more of sucrose that is solute
and then you define the volume of
solution in leader that is 0.35 in it
and you get that correct so this guy is
involved in this guy's in liter so the
answer is volt one liter then he has
something called molality this guy is
nothing but number of moles of solute
present in 1 kg of solvent please note
in earlier case it was one liter of
solvent here it is 1 kg of solvent right
the concept is same that is noted by
small M so I will write small M is
nothing but most of solute in
kg of solvent this guy is in moles and
this guy is in kg the unit will be more
per case in case of molarity it was mole
per liter correct so in case of molarity
capital M it was mole of solute by
liters of solution correct this guy was
mole per liters
this is mole per liter so important this
guy is both Berkut this Sigma example
for this simple example fool kg of sugar
dissolve and 350 religion of water
what is the molarity here density of
water in school so again same thing
first thing we'll do find them all molar
mass of c12 h22 o11 will come out to be
12 into 2 n plus hydrogen 12 well into
111 oxygen 11 to 16 that is 342 grams
per mole so 342 grams of sucrose go to
one wood but we have 4 grams so 4 grams
of sucrose same thing I am doing here is
equal to 4 y 342 moles that is point
zero one Lancelin no this is my board
now I have to find small n morality it
is nothing but mole of sugar by kg of
water so density is equal to mass by
volume I know because I am talking about
mass of water so the water will be then
C 2 volume what is density densities
0.97 gram per milliliter gram per
milliliter so I have to multiply this
guy with milliliter 350 million so what
you will get in
is in Graham right because gram per
milliliter into something in later
millimeter as will you get something in
Graham so you have to multiply this guy
with 10 to the power minus so you get
something in kg hope you understand this
so what you get here is 0.34 one case so
I got the mass of water so I can say
small m whether it is nothing but mole
of c12h22o11 by mass of water this guy
will be in mole this there will be in
kitchen right this becomes I have zero
point one one seven and this guy is in
kg is point three for 1kg so you solve
this digadz point zero three for voltage
and that is my molarity easy only so the
only difference is here instead of in
molarity two as volume of water here it
is mass of watering it is a common
question we have to find the mass of
sodium acetate required to make 50 ml of
0.375
molar equus solution whether molar mass
of sodium as it is this so if you see in
this case i have to find the mass of
this guy required to make this molar
solution
so then i say it's 0.375 molar solution
this guy means i have point three seven
five moles of ch3 Co la in 1000 ml of
one correct this is what means or I'm
talking about 5 m level of water you
divide everything by 2 that means I have
point one eight seven five mole of ch3
Co na in 500 ml of h2
correct so now I have to find how much
gram this guy is I know that eighty two
point zero two four five gram one mole
is equal to this much laughs you just
much I know
one more is well this much time so
0.1875 mole will be equal to 0.1875 into
this much adieu two point zero two four
five grams and this if you solve you get
fifteen point three that is mask hope
you understand see point three seven
five molar solution means 0.375 moloch
sorry mess it in 100 ml of water since
we have five in MLM bottom that means we
divide everything by 2 that is 0.1875
mole of sodium acetate in 5 ml of water
so point nami to find the question says
you define how many grams of sodium
acetate in this mesh mode so we have
told that one mole is equal to 82 point
zero two five grams so 0.875 187 500
will be
0.1875 into eighty two point zero to PI
gram and thence come out to be 150
fifteen point three thank you visit exam
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