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ChemistryNumerical+3Question 34 of 54

JEE Advanced 2019 · Paper 1, Chemistry question 16

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 and worked solution

Official answer: 0.97

Moles of solute. The relative lowering of vapour pressure is the mole fraction of solute:
10650=n2n1+n2,n1=3978=0.5  ⟹  n2=0.564=7.81×10−3 mol.\frac{10}{650}=\frac{n_2}{n_1+n_2},\qquad n_1=\frac{39}{78}=0.5\implies n_2=\frac{0.5}{64}=7.81\times10^{-3}\text{ mol}.

Freezing point.
m=7.81×10−30.039=0.200 mol kg−1,ΔTf=5.12×0.200=1.02 K.m=\frac{7.81\times10^{-3}}{0.039}=0.200\ \mathrm{mol\,kg^{-1}},\qquad\Delta T_f=5.12\times0.200=1.02\text{ K}.
The dilute approximation n2≈n1⋅10650n_2\approx n_1\cdot\frac{10}{650} gives 1.01 K.

Answer: 1.02 (key: 0.97 to 1.06)

Topics:Colligative propertiesRaoult's law and vapour pressure
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