00:03
Here in this problem, given boiling point of a solution is 106 .5 degrees centigrade.
00:10
We have to find out the freezing point of that solution.
00:14
Now, delta t b is, that is the elevation of boiling point, delta t b is t b boiling point of the solution minus t b of water, boiling point of the solvent here boiling point of the solution is 106 .5 degrees centigrade and boiling point of water is 100 degrees centigrade.
00:46
So delta t b the elevation of boiling point is 6 .5 degree centigrade.
00:53
Now we know you can write this formula elevation of boiling point is kb that moral elevation boiling point elevation constant times molality delta t b we know 6 .5 and kb for water 0 .512 degrees 8 decade per molal and here molality so from here we can find the molality of the solution 6 .5 divided by 0 .512 and we get 12 .7 now we have to find out the freezing point of the solution.
01:34
Let's write the formula, depression of freezing point, delta tf is equal to kf moral freezing point, depression or freezing point constant, times molality.
01:48
Now, we know the kf is one, k for water, 1 .86 times molality 12 .7.
01:56
So delta t f that is the depression of freezing point is 23 .6 degree centigrade...