JEE Main 2026 — Jan 24, Shift 1 — Chemistry solution

JEE Main2026Jan 24, Shift 1Chemistry

'W' g of a non-volatile electrolyte solid solute of molar mass 'M' g 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 $K_{b}$ for water is 0.52 K kg mol $^{- 1}$ , then the mole fraction of the electrolyte solute ( $x_{2}$ ) in the solution can be expressed as

(Given : density of water = 1 g/mL and boiling point of water = 373 K)

A
$\frac{1.3}{8} \times \frac{W}{M}$
B
$\frac{16}{2.6} \times \frac{W}{M}$
C
$\frac{2.6}{16} \times \frac{M}{W}$
D
$\frac{1.3}{8} \times \frac{M}{W}$

Solution

Step 1 — find van't Hoff factor $i$ from $Δ T_{b}$ :

$Δ T_{b} = 375 - 373 = 2$ K, $m = \frac{W / M}{0.1} = \frac{10 W}{M}$ mol/kg.

$Δ T_{b} = i \cdot K_{b} \cdot m \Rightarrow 2 = i \cdot 0.52 \cdot \frac{10 W}{M}$

$\Rightarrow i = \frac{2 M}{5.2 W} = \frac{1}{2.6} \cdot \frac{M}{W}$ .

Step 2 — relate $Δ P / P^{\circ}$ to mole fraction with the same $i$ :

$\frac{Δ P}{P ^{\circ}} = i \cdot x_{solute} \Rightarrow \frac{40}{640} = i \cdot x_{2}$

$\Rightarrow x_{2} = \frac{1}{16 i} = \frac{1}{16} \cdot \frac{2.6 W}{M} = \frac{2.6}{16} \cdot \frac{W}{M} = \frac{1.3}{8} \cdot \frac{W}{M}$ .

← Back to full paper

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →