JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Solutions

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At room temperature, a dilute solution of urea is prepared by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be: (molar mass of urea

= 60 g mol^{- 1}

)

(A)0.027 mmHg
(B)0.031 mmHg
(C)0.028 mmHg
(D)0.017 mmHg✓ Correct

Answer: (D)

Solution

By Raoult's law the lowering of vapour pressure is

Δ p = p^{0} x_{so l u t e}

. Urea gives

\frac{0.60}{60} = 0.01

mol and water gives

\frac{360}{18} = 20

mol, so the mole fraction of urea is

\frac{0.01}{0.01 + 20} = \frac{0.01}{20.01}

. Hence

Δ p = 35 \times \frac{0.01}{20.01} \approx 0.017

mmHg.

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option

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