Sea water contains $29.25 \% \mathrm{NaCl}$ and $19 \% \mathrm{MgCl}_{2}$ by weight of solution. The normal boiling point of the sea water is ____ $^{\circ} \mathrm{C}$ (Nearest integer) Assume $100 \%$ ionization for both NaCl and $\mathrm{MgCl}_{2}$ Given: $\mathrm{K}_{\mathrm{b}}(\mathrm{H}_{2} \mathrm{O})=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ Molar mass of NaCl and $\mathrm{MgCl}_{2}$ is 58.5 and 95 g $\mathrm{mol}^{-1}$ respectively.

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Solutions

Sea water contains

29.25% NaCl

and

19% MgCl_{2}

by weight of solution. The normal boiling point of the sea water is ____

^{\circ} C

(Nearest integer)

Assume

100%

ionization for both NaCl and

MgCl_{2}

Given:

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

Molar mass of NaCl and

MgCl_{2}

is 58.5 and 95 g

mol^{- 1}

respectively.

Answer: 116

Solution

In 100 g solution: 29.25 g NaCl, 19 g MgCl₂, solvent = 100 - 48.25 = 51.75 g = 0.05175 kg.

NaCl ionizes to give i = 2 (Na⁺ + Cl⁻); MgCl₂ gives i = 3 (Mg²⁺ + 2Cl⁻).

Molality contributions:

NaCl: $m_{1} = \frac{29.25}{58.5 \times 0.05175}$
MgCl₂: $m_{2} = \frac{19}{95 \times 0.05175}$

Boiling point elevation:

Δ T_{b} = K_{b} (i_{1} m_{1} + i_{2} m_{2}) = 0.52 [\frac{2 \times 29.25}{58.5 \times 0.05175} + \frac{3 \times 19}{95 \times 0.05175}]

Δ T_{b} = 0.52 [\frac{58.5}{3.027} + \frac{57}{4.916}] = 0.52 [19.33 + 11.60] = 0.52 \times 30.93 \approx 16.08

Boiling point = 100 + 16.08 ≈ 116°C.

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Integer answer

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JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Solutions

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