Question

Four solutions of ${\mathrm{K}}_2{\mathrm{SO}}_4$ with the following concentration $0.1\mathrm{m},0.01\mathrm{m},0.001\mathrm{m}$ and $0.0001\mathrm{m}$ are available. The maximum value of van 't Hoff factor$\left(\mathrm{i}\right)$ corresponds to

• A. $0.1\mathrm{m}$ Solution
• B. $0.01\mathrm{m}$ Solution
• C. $0.001\mathrm{m}$ Solution
• D. $0.0001\mathrm{m}$ Solution

Answer 1

The correct option is D

$0.0001\mathrm{m}$ Solution

An explanation for the correct answer;

Van 't Hoff factor;

It is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

As we know, the freezing point depression equation is $∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\mathrm{mi}$

Where,

${\mathrm{k}}_{\mathrm{f}}$ = cryoscopic factor

$\mathrm{m}$=molality

$\mathrm{i}$ = Van’t Hoff factor

By rearranging this equation we get,

$\mathrm{i}=\frac{∆{\mathrm{T}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{f}}\mathrm{m}}$

From this equation, we conclude that Van’t Hoff factor ‘$\mathrm{i}$’ is inversely proportional to molality $\mathrm{m}$.

That is, $\left(\mathrm{i}\propto \frac{1}{\mathrm{m}}\right)$

Therefore the solution with a low concentration will have a maximum van 't Hoff factor$\left(\mathrm{i}\right)$.

So, the solution of ${\mathrm{K}}_2{\mathrm{SO}}_4$ with concentration $0.0001\mathrm{m}$ is having the maximum van 't Hoff factor$\left(\mathrm{i}\right)$.

Hence the option $\mathrm{D}$ is correct.