00:03
Here in this problem, given freezing point of acetic acid is 16 .6 degrees celsius, mass of solute, which is diphenile ether, that is 32 gram, mass of solvent, that is acetic acid, 192 .9 gram.
00:19
Kf for acetic acid given, we have to find out the freezing point of the solution.
00:26
Now, let's find out moles of solute, that is diphenile ether.
00:35
Most of solute the mass given here 32 gram so let's write that 32 gram divided by molar mass of diphenyl ether which is 170 .2 gram per mole and we get 32 divided by 170 .2 more now here the mass of solvent is 192 .9 gram so mass of solvent mass of solvent 192 .9 gram so we'll convert it into kilogram 192 .9 times 10 to the power minus 3 kilogram since one gram is 10 to the power minus 3 kilogram now we can find out the molal of the solution.
01:41
Morality of the solution is moles of solute, which is this one.
01:48
So let's write 32 divided by 170 .2 moles divided by mass of solvent in kilogram, this one.
02:02
192 .9 times 10 to the per negative 3 kilogram.
02:08
And this is 0 .947 .9.
02:16
Mollal.
02:17
Now, let's write the formula, delta t .a.
02:23
Friesing point depression is equal to kf times, same.
02:29
A molality, kf, depression, depression.
02:32
Increasing point constant since it is a non -electrolite so here a vent of factor is one so we are just writing k f times m now k f given here for acetic acid the solvent this one 3 .90 degree centigrade per molal and molality we just calculated here 0 .0...