JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Solutions

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On dissolving 0.5 g of a non-volatile non-ionic solute to 39 g of benzene, its vapor pressure decreases from 650 mm Hg to 640 mm Hg . The depression of freezing point of benzene (in K) upon addition of the solute is ____
(Given data : Molar mass and the molal freezing point depression constant of benzene are

78 g mol^{- 1}

and

5.12 K kg mol^{- 1}

, respectively)

Answer: 1.015

Solution

Using Raoult's law for vapor pressure lowering:

\frac{P ^{\circ} - P _{s}}{P ^{\circ}} = \frac{n _{solute}}{n _{solute} + n _{solvent}}

Moles of benzene:

n_{solvent} = \frac{39}{78} = 0.5 mol

\frac{650 - 640}{650} = \frac{n _{solute}}{n _{solute} + 0.5}

\frac{10}{650} = \frac{n _{solute}}{n _{solute} + 0.5}

Since

n_{solute} ≪ 0.5

n_{solute} \approx \frac{10}{650} \times 0.5 = \frac{5}{650} = \frac{1}{130} mol

Molality:

m = \frac{n _{solute}}{mass of solvent (kg)} = \frac{1/130}{0.039} = \frac{1000}{130 \times 39} = \frac{1000}{5070} \approx 0.197 mol/kg

Freezing point depression:

Δ T_{f} = K_{f} \times m = 5.12 \times 0.197 \approx 1.01 K

The source key accepts 0.97 to 1.06 K.

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Numerical answer

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JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Solutions

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