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 mmHg, lowering of vapour pressure will be (molar mass of urea = 60 g mol

Engineering

Chemistry

Relative Lowering in Vapour Pressure

Raoult Law and Vapour Pressure

Question

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 mmHg, lowering of vapour pressure will be

(molar mass of urea = 60 g mol^–1):-

0.028 mmHg

0.031 mmHg

0.017 mmHg

0.027 mmHg

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lowering of vapour pressure = P⁰ – P = P⁰.x_solute

$∴ Δp = 35 \times \frac{0 .6 / 60}{\frac{0 .6}{60} + \frac{360}{18}} = 35 \times \frac{. 01}{. 01 + 20} = 35 \times \frac{. 01}{20 .01} = 0 .17 mm Hg$