JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

' W ' g of a non-volatile electrolyte solid solute of molar mass ' M ' $\mathrm{g} \mathrm{mol}^{-1}$ when dissolved in 100 mL water, decreases vapour pressure of water from 640 mm Hg to 600 mm Hg. If aqueous solution of the electrolyte boils at 375 K and $\mathrm{K}_{\mathrm{b}}$ for water is $0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ , then the mole fraction of the electrolyte solute $\left(x_{2}\right)$ in the solution can be expressed as

(Given : density of water $=1 \mathrm{~g} / \mathrm{mL}$ and boiling point of water $=373 \mathrm{~K}$ )

Official previous-year question

Held on 24 Jan 2026 · Verified 6 Jul 2026.

Options

A
$\frac{2 \cdot 6}{16} \times \frac{\mathrm{M}}{\mathrm{W}}$
B
$\frac{16}{2 \cdot 6} \times \frac{\mathrm{W}}{\mathrm{M}}$
C
$\frac{1.3}{8} \times \frac{\mathrm{M}}{\mathrm{W}}$
D
$\frac{1 \cdot 3}{8} \times \frac{\mathrm{W}}{\mathrm{M}}$

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Solution

Given the vapour pressure of pure water $P^o = 640$ mm Hg and vapour pressure of solution $P_s = 600$ mm Hg.

According to Raoult's law for an electrolyte solution, the relative lowering of vapour pressure is given by $\frac{P^o - P_s}{P^o} = i \cdot x_2$ , where $i$ is the van't Hoff factor and $x_2$ is the mole fraction of the solute.

Substituting the values: $\frac{640 - 600}{640} = i \cdot x_2$

$\frac{40}{640} = i \cdot x_2 \implies i \cdot x_2 = \frac{1}{16}$ ... (1)

The elevation in boiling point is given by $\Delta T_b = i \cdot K_b \cdot m$ , where $m$ is the molality.

Given $T_b = 375$ K and $T_b^o = 373$ K, so $\Delta T_b = 375 - 373 = 2$ K.

Molality $m = \frac{W}{M \times 100} \times 1000 = \frac{10W}{M}$ (since mass of water = $100$ mL $\times 1$ g/mL = $100$ g).

Substituting in the elevation formula: $2 = i \cdot 0.52 \cdot \frac{10W}{M}$

$i = \frac{2M}{5.2W} = \frac{20M}{52W} = \frac{5M}{13W}$ ... (2)

Substitute the value of $i$ from (2) into (1):

$\frac{5M}{13W} \cdot x_2 = \frac{1}{16}$

$x_2 = \frac{1}{16} \cdot \frac{13W}{5M} = \frac{13}{80} \cdot \frac{W}{M}$

This can be rewritten as $x_2 = \frac{1.3}{8} \times \frac{W}{M}$ .

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Options

$\frac{2 \cdot 6}{16} \times \frac{\mathrm{M}}{\mathrm{W}}$

$\frac{16}{2 \cdot 6} \times \frac{\mathrm{W}}{\mathrm{M}}$

$\frac{1.3}{8} \times \frac{\mathrm{M}}{\mathrm{W}}$

$\frac{1 \cdot 3}{8} \times \frac{\mathrm{W}}{\mathrm{M}}$