A solution containing $2.5 \times 10^{-3} \mathrm{~kg}$ of a solute dissolved in $75 \times 10^{-3} \mathrm{~kg}$ of water boils at 373.535 K. The molar mass of the solute is ______ $\mathrm{g~mol}^{-1}$. [nearest integer] (Given: $\mathrm{K}_{\mathrm{b}}\left(\mathrm{H}_{2}\mathrm{O}\right)=0.52 \mathrm{~K~Kg~mol}^{-1}$, boiling point of water $=373.15 \mathrm{~K}$)

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Solutions

A solution containing

2.5 \times 1 0^{- 3} kg

of a solute dissolved in

75 \times 1 0^{- 3} kg

of water boils at 373.535 K. The molar mass of the solute is ______

g mol^{- 1}

. [nearest integer]
(Given:

K_{b} (H_{2} O) = 0.52 K Kg mol^{- 1}

, boiling point of water

= 373.15 K

)

Answer: 45

Solution

The elevation in boiling point is given by

Δ T_{B} = K_{b} \times m

, where

m

is molality.

Δ T_{B} = 373.535 - 373.15 = 0.385

K

Molality

m = \frac{moles of solute}{kg of solvent} = \frac{w / M}{W _{solvent}}

Substituting:

0.385 = 0.52 \times \frac{2.5}{M} \times \frac{1000}{75}

Solving for

M

M = \frac{0.52 \times 2.5 \times 1000}{0.385 \times 75} = \frac{1300}{28.875} \approx 45

g mol

^{- 1}

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Integer answer

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