JEE Main 2025 — Solutions Question with Solution

Freezing Point Depression:

$\Delta T_f=i\times K_f\times m$

Given:

$\Delta T_f=0.2\text{°C},K_f=1.8{\text{K kg mol}}^{-1},m=0.1\text{m}$

Substituting the given values:

$0.2=i\times 1.8\times 0.1$

Solving for $i$:

$i=\frac{0.2}{1.8\times 0.1}=\frac{20}{18}=\frac{10}{9}$

Degree of Dissociation ($\alpha$):

For the reaction $\mathrm{HA}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{A}}^{-}$:

$i=1+\alpha$

Given $i=\frac{10}{9}$:

$\frac{10}{9}=1+\alpha$

$\alpha =\frac{1}{9}$

Dissociation Constant ($K_{eq}$):

${\mathrm{K}}_{eq}=\frac{[H^{+}][A^{-}]}{[HA]}$

At equilibrium:

$[H^{+}]=[A^{-}]=\alpha \times C=\frac{1}{9}\times 0.1$

$[HA]=0.1\times (1-\alpha )=0.1\times \left(1-\frac{1}{9}\right)$

Substituting these into $K_{eq}$:

${\mathrm{K}}_{eq}=\frac{(0.1\times \frac{1}{9}{)}^2}{0.1\times \left(1-\frac{1}{9}\right)}$

Simplifying:

${\mathrm{K}}_{eq}=\frac{0.1\times \left(\frac{1}{81}\right)}{0.1\times \frac{8}{9}}=\frac{1}{720}$

Therefore:

${\mathrm{K}}_{eq}=1.38\times 10^{-3}$