Question

What is molal depression constant ? An aqueous solution freezes at −0.385 $ ^o C $ if $ {K_f} $ is 3.85 K kg mol−1 and $ {K_b} $ ​=0.712 K kg mol−1. What is the elevation in boiling point ?

Answer 1

Hint: Initially you must be aware of the term molal depression constant. After knowing the terminology your prime f $ ^o C $ us must be towards what is to be found. Here everything is given just $ \Delta {T_b} $ is just to be found. Moreover, one should be careful with the units.

Complete step by step solution:
The depression which is observed in freezing point in a solution in which 1 gm mole of solute is dissolved in 1000 gm of solvent is termed as molal depression constant. It is represented by $ {K_f} $ ​.
In other words, “The freezing point depression ( $ \Delta {T_f} $ ) of solutions of nonelectrolytes has been found to be equal to the molality (m) of the solute times a proportionality constant called the molal freezing point depression constant , $ {K_f} $ .”
Given:
Freezing point of solution ( $ T_2 $ ) = −0.385 $ ^o C $
 $ T_1 $ = 0 $ ^o C $
 $ {K_f} $ ​=3.85 K kg mol−1
 $ {K_b} $ =0.712 K kg mol−1
We know that elevation in boiling point is given by
 $ \Delta {T_b} = {K_b} \times m $
 $ \Delta {T_f} = {T_1} - {T_2} $
= 0-(-0.385)
= 0.385 $ ^o C $
Now molality (It is a property of a solution and is defined as the number of moles of solute per kilogram of solvent.):
 $ m = \dfrac{{\Delta {T_f}}}{{{K_f}}} $
 $ m = \dfrac{{0.385}}{{3.85}} $
= 0.1 M
Elevation in boiling point= $ \Delta {T_b} $
 $ \Delta {T_b} = {K_b} \times m $
=0.712 X 0.1
=0.0712 $ ^o C $
Hence elevation in boiling point is 0.0712 $ ^o C $ .

Note:
One should take care to not use the mass of solution. Often, the matter can be that the question. They can bring change in temperature and also the proportion constant, and you have to realize the molality at first so as to urge your final answer. The solute, so as for it to exert any modification on colligative properties, should fulfil two conditions. First, it should not contribute to the vapour pressure of the solution, and second, it should stay suspended within the answer even throughout part changes. as a result of the solvent is not any longer pure with the addition of solutes, we are able to say that the chemical potential of the solvent is lower.