0:00
Hello.
00:01
In this question here we are given this table.
00:04
So here first we need to calculate here that is the freezing point of depression.
00:11
That is delta t .f.
00:14
So therefore here that is equal to this temperature of pure loric acid.
00:21
We know that it is equal to temperature of this pure loric acid minus temperature of this loric acid with benzoic acid so here as we know that that is the temperature here so here it is as it is already given here this freezing point of depression here that is 44 minus 39 it is 5 degree 44 minus 39 it will be 5 degrees celsius for loric acid here and this is 44 minus 37 that will be 7 degrees celsius.
01:05
Now here we can find out this molality.
01:09
So here molality it is equal to we know that that is raising point of depression divided by this kf value and kf value for loric acid it is 3 .90.
01:26
So therefore here for for loric acid and benzoic acid.
01:34
So therefore here on putting the value we will have that is 5 is the temperature so 5 divided by 39 it is there so therefore on solving this we will have this value that is 1 .28 here we will have that is 1 .28 2 it is there similarly we can find out for this loric acid with 0 .5 gram of benzoyc acid so this will become so here this will become 6 divided by 6 divided by 3 .9 this is 3 .9.
02:16
3 .9 it is equal to and hence on solving we will have 1 .538 it is there 1 .538.
02:24
Now, moles of benzoic acid.
02:27
As we know that, that is for this moles, it is equal to here mass.
02:36
It is given here.
02:38
That is the molality.
02:40
Molality multiplied by mass of loric acid in kilogram.
02:45
So therefore here we have this number of moles.
02:50
If we see for the, let us suppose this is for column 8, this is for column 8.
02:54
So that we do not need to write down again and again.
02:57
This chloric acid plus benzodia acid etc.
03:00
So here this molality is 1 .282 multiply by...