Video: Abnormal Molar Masses Video Lecture - NEET

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no more fear from exam let's understand
a very critical topic on abnormal
monomers before we tell you what does
that let's see this equation example in
cystic ionic compound and ionic Aborn
journey dissolves in water and it
dissociate into cations and anions for
example I have case here the moment you
put in water this becomes K plus ions
elements on assembly you see I have put
K sillens water and KCl has is broken
into K plus ion and cl- ions right that
means if I put one particle of KCl that
actually becomes two particle and I have
seen that all the qualitative property
we have discussed it is increasing in
vapour pressure or decrease in melting
point boiling point is kind of staff's
osmotic pressure these qualitative
properties depends only on the number of
particles and that is why they are all
calling the problem now in such case if
you think if we talk about the ionic
compounds the moment you put in a quest
solution and we do it lot because water
is universal solvent the number of
particles doubles right and as I told
that the colligative properties depends
only the number of particles the
expectation behavior is the boiling
point which we expect decreasing or the
increase in volume 1 which we expect
that should be double why see because I
told delta T model is equal to KF into M
and this is nothing but the number of
particles now in case of ionic solution
and if you are using the any component
water ionic equation then the number of
particle is double number of particle is
defined so the change in temperature
should also be double correct so the
boiling point would also be Devin but
the dissociation is not foreign to us
you see you must have seen this thing
kppm right all this K constant so the
dissociation is not 100% maybe it is 90%
or 80% I don't know it depends on the
factors right so the experimental
determine value the experimental value
of this will be different from this is
calculated so the calculated value of an
experimental value will not match they
will differ also if you take another
example if you take the ethanoic acid
this is mine I gather so you're throwing
acid when you put in benzene in benzene
that we have seen Benjen is my solvent
and it - it is my solute
right so what happened is this it no I
guess it died Verizon right due to - you
want they time rise so if you see in
that case the number of molecules is
half because if there were 10 molecules
of ethanoic acid another day it became 5
minutes 10 we came five minutes right
because two became one unit like that
so expectation is that again here in
this case the boiling point or the
change in boiling point or the freezing
point what we are expecting a machine in
pressure it should be half of the
expected value of the carotids value
right but that doesn't also happen
properly when it's not it exactly half
it may be that it is 1.7 times 1.6 times
so experimental value in the capital
value is not same always right so such
molar mass that is either lower or
higher than they expected or the normal
value is called abnormal molar mass see
you see the the molarity is actually
either higher or lower than the expected
order to one to one to take care of this
vant off key because of a factor call
van
tough fat so in 1880 this guy ranch off
he gave a factor he introduced a factor
I he due to the factor I and this is
called vant Hoff factor right it account
for the extent of this association or
Association so it account for the extent
of association or this by so see agent
for example if I have KCl it will become
K plus C minus so this guy will take
care of that k plus then C ever so it
will take care of the association out
this association in the case of my as I
just explained acetic acid or if you
know I guess it is actually become half
amazing correct so this value of I is
nothing but total number of particles
after my this a sufficient or
association or divided by total number
of particle before this association
organization example in this case of kcl
sample in this case of KCl it be it
gives K plus and Cl minus let's suppose
that T is equal to zero I had one
particle and there were zero particles
here T is equal to P when equilibrium is
reached this becomes 1 minus SX particle
this person this becomes X system X
right so total number of particles after
that this position will be how much 1
minus X plus
X plus X there is something but one plus
X and total number of particles before
this thing started is one so I will be
in this case 1 plus X by one hope you
understand total number of particles
after dissolution dissolution because
not all the casein will disassociate
into k plus and Cl minus only X of these
KCl will dis so she didn't do k plus NC
Mal's right total my body has after this
position will be 1 plus X total number
of particles initially is 1 so the Y
value is 1 plus X Y correct and it is
also generally observed colligative
property by calculator calculator
problem so when there is for example the
value of I for a consolidation is almost
2 one must is almost 2 I for this case
if I get it I is almost 1 by 2 almost ok
so with this vant Hoff factor and this
sense is a very important factor and
other equations other scientists also
realized it so they changed other
equations so the equation got changed
the writer factor were think read in the
equation which we have just learned
right example change in PV was earlier
KB into M now it is I into kV into M so
nearly change in a freezing point is
nothing but I into hairpin - mmm same
thing P 1 becomes P 1 naught into X 1
earlier so here I also king right
similarly for osmosis pressure was n RT
by V so here also I came in the picture
so now all these equations I can look
please note these are the final equation
after the vant Hoff factor is taken into
account
now this equation I came into picture
when talk about change boiling point
change in freezing point in pressure
right we talked about change in pressure
actually Delta P I is nothing but P I
naught into n2 by n1 this we have seen
per target solution and this here also I
came into picture
some of the rank top factor is given
right four point one molar solution if
you see the value of I for NaCl is one
point eight seven if you dilute it the
value of I increase to dilute it further
the value of I for increase right
similarly for KCl potassium chloride the
value of I 4.1 and solution is one point
eight five
if you diluted the value increased to
almost 2 so the value of I also depends
on the molarity of the solution
I will take some numerical all this the
question says two gram of benzoic acid
dissolved in benzene this is my benzene
here my red one is my benzoic acid right
so hey this is my to ram this is
dissolved in bins in the benzene 25 gram
benzene correct it shows a molar
depression freezing point is delta TF it
shows us one point six two K right here
okay and the KF value of benzene is
given four point nine k kg per mole you
define the percentage dissipation of
acid if it forms dimer in the solution
so it is told that C 6 H 5 CE o H will
form dimer
in this solution we have to find a
percentage association actually so let's
suppose that T is equal to zero it was
one it was 0 at T is equal to TF Librium
becomes 1 by X this becomes X by 2 I is
what I is what at final product it is 1
minus X plus X total concentration at T
is equal to T is 1 1 minus X plus X by 2
there is nothing but 1 minus X by 2 and
what is the concentration at initial
please 1 a so I will become this time
correct so now the formula I know is
delta TF is nothing but I into here into
M let us put the value here Delta TF is
how much 1 point 6 2 K is nothing but I
is given 1 minus expected correct and KF
is given 4 point 9 k ilg per unit book
into M M is what M is my molality of
benzoic acid how much is the gram of
benzoic acid to Graham what is the molar
mass of benzoic acid it is 122 grams and
one right into mass of solvent in kg is
25 man this becomes 1 by 25 you comes mm
actually you may clean kitchen correct
when you solve this equation you will
get X as nothing but 0.99 2 that is what
will get X is 0.99 - what does it mean
this means that degree of Association of
Bin's
benzoic acid is 90
9.2 percent this because 0.99 to x 800
get percentage this is ninety nine point
two is a person is a degree of
especially Nam benzoic acid it easy just
they found the value of I in in
percentages or in the terms of a
sufficient and then I could get the
value of x and that is nothing but my
association constant the question says
point six ml of acetic acid having
density of one point zero six gram per
ml the dissolved in one liter of water
this is one liter water the decoration
is freezing point was observed to be
point zero to five and see as it is
delta T a is point zero two zero five
I told degree Celsius degree Kelvin is
all saying I've explained this in
another numericals why I can write this
as K need to find the vent of factor and
the Association cost
so let's find meant affected first I
know Delta TF is nothing but I into KF
into M correct so first find em I easy
to find K is I think known to us K for
water is a one point eight six key kg
per would this is something known to us
this is known to us let's find em first
modality of s the gas is nothing but
mass of acetic acid by molar mass of
acetic acid into kg of solvent and is kg
of water water e kg what is the mass of
esta casa volume which is given density
is given density is equal to mass by
volume mass is equal to density into
volume density is one point zero
gram per milliliter and bulb and volume
is 0.6 mm into 0.6 mm wait this whole
thing in 226 ml by molar mass of acetic
acid
I know a 60 gram per mole right we just
find 12 carbon plus 3 hydrogen 12 carbon
16 16 32 is into oxygen one hydrogen
atom 60 gram per mole into ma mass of
water 1 liter of water is almost 1 kg of
water almost this tea is going to rate
suppose one gram right so there is
almost 1 minute later we later canceled
gram gram cancel you solve this you get
almost 0.1 0-6
would mole per kg because kg H okay so
let's put the value Delta TF is what
point 0 to 0-5 Kelvin is equal to I into
this value k 1.86
e kg per mole into this value point 0 1
0 6 mole per kg right biscuits
kg leaves cancel Kelvin Kelvin gets
cancel what I get is I is one point zero
four one
this is unitless so I have found the
value of I now let's find the value of
sufficient constant so what I know is I
is equal to one point zero correct so
let's see the equation we have ch3cooh
it will form h plus and ch3co - water so
T is equal to zero let's assume the
concentration is n 0 it is 0 T is equal
to Q what will happen this becomes n
one minus X this becomes the next this
becomes an X I is what concentration at
equilibrium that is n into 1 minus X
plus NX plus NX
if you do that that comes out to be into
1 minus X plus X plus X divided by
concentration at T is equal to 0 it is n
correct you solve it you get 1 plus X so
Y is nothing but 1 point 0 4 1 is
nothing but 1 plus X you saw this you
get X is nothing but point 0 so degree
of dissipation of acetic acid is point 0
4 now I have to find K K for this right
K for this equation K is an event
concentration of H+ in the concentration
of ch3co - by concentration of ch3cooh
correct concentration of h+ is nothing
but n x you know that concentration of
CSC so - and also an X and this is ne
into 1 minus X and then cancel now you
put the values here what is the value of
n NX square n we have seen the molarity
was my name 0 1 6 X square X is what
point 0 4 1 square you just want a value
of x divided by 1 minus point zero 1 you
solve this you get 1 point 8 6 into 10
to power minus that is the dissociation
constant for this
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