A solution containing 2 g of a non-volatile solute in 20 g of water boils at 373.52 K. The molecular mass of the solute is ______ $\mathrm{g} \mathrm{mol}^{-1}$. (Nearest integer) Given, water boils at $373 \mathrm{~K}, \mathrm{~K}_{\mathrm{b}}$ for water $=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Solutions

A solution containing 2 g of a non-volatile solute in 20 g of water boils at 373.52 K. The molecular mass of the solute is ______

g mol^{- 1}

. (Nearest integer)

Given, water boils at

373 K, K_{b}

for water

= 0.52 K kg mol^{- 1}

Answer: 100

Solution

The elevation of the boiling point is

Δ T_{b} = 373.52 - 373 = 0.52 K

. Using

Δ T_{b} = K_{b} m

with molality

m = \frac{2}{M} \times \frac{1000}{20}

, the relation becomes

0.52 = 0.52 \times \frac{2 \times 1000}{20 M}

. Solving,

M = \frac{0.52 \times 100}{0.52} = 100

. The molar mass of the solute is

100 g mo l^{- 1}

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Integer answer

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