At room temperature, a dilute solution of urea is prepared by dissolving 0.60 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\(^{-1}\))
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- 0.031 mmHg
- 0.017 mmHg
- 0.028 mmHg
- 0.027 mmHg
The Correct Option is B
Solution and Explanation
Moles of water = 360 g / 18 g mol\(^{-1}\) = 20 mol.
Relative lowering: \(\frac{P^0-P_s}{P^0}=x_{urea}=\frac{0.01}{0.01+20}=\frac{0.01}{20.01}\approx 5.0\times10^{-4}\).
\(\Delta P = 35\times 5.0\times10^{-4}\approx 0.0175\) mm Hg, i.e. about 0.017 mmHg.
Answer: (2) 0.017 mmHg
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