Question

What is the definition of molal depression constant?
An aqueous solution freezes at $- {0.385^0}C{\text{ if }}{{\text{K}}_f} = 3.85K{\text{ kgmo}}{{\text{l}}^{ - 1}}{\text{ and }}{{\text{K}}_b} = 0.712kg{\text{ mo}}{{\text{l}}^{ - 1}}$ find will be the elevation in boiling point?

Answer 1

Hint – In this question use the basics of molal depression constant which is denoted by ${K_f}$ that it is the depression in freezing point for a particular quantity of solute dissolved in a particular amount of solvent. For the second part use the direct formula that is $\Delta {T_b} = {K_b} \times m$, where $m = \dfrac{{\Delta {T_f}}}{{{K_f}}}$, this will help getting the answer.

Complete answer:
Molal depression constant:
The depression in freezing point in a solution in which 1 gm mole of solute dissolved in 1000 gm of solvent is known as molal depression constant represented by ${K_f}$.
Numerical:
Given value, Freezing point of solution = ${T_2} = - {0.385^0}C$
Molal depression constant, ${K_f} = 3.85$ K kg mol-1
${K_b} = 0.712$ K kg mol-1
We have to find out elevation in boiling point, $\Delta {T_b} =$ ?
Then, $\Delta {T_f} = {T_1} - {T_2}$
= 0 – (- 0.385) = ${0.385^0}C$
Now molality (m) is the ratio of $\dfrac{{\Delta {T_f}}}{{{K_f}}}$
$m = \dfrac{{\Delta {T_f}}}{{{K_f}}} = \dfrac{{0.385}}{{3.85}} = 0.1$
Therefore elevation in boiling point is the multiplication of ${K_b}$ and molality.
$\Rightarrow \Delta {T_b} = {K_b} \times m$

Now substitute the values we have,
$\Rightarrow \Delta {T_b} = 0.7120 \times 0.1 = {0.0712^0}C$
So this is the required answer.

Note – Elevation in boiling points describes the phenomena that the boiling point of a liquid (or any solvent) will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent. This actually happens when a non-volatile solute is added to a pure solvent.