At room temperature, a dilute solution of urea is prepared by dissolving $0.60 g$ of urea in 360g of water. If the vapour pressure of pure water at this temperature is $35 m m H g$ . lowering of vapour pressure will be: (molar mass of urea $= 60 g m o l^{- 1}$ )

0.027 m m H g

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0.028 m m H g

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0.017 m m H g

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0.031 m m H g

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Solution

The correct option is B $0.017 m m H g$
Solution:- (C) $0.017 m m H g$

Lowering of vapour pressure $= P^{0} - p = p^{0} . x_{s o l u t e}$

$∴ Δ P = 35 \times \frac{0.6 / 60}{\frac{0.6}{60} + \frac{360}{18}}$

$= 35 \times \frac{.01}{.01 + 20} = 35 \times \frac{.01}{20.01}$

$= 0.17 m m H g$

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At room temperature, a dilute solution of urea is prepared by dissolving 0.60g of urea in 360g 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 gmol−1)

The correct option is B 0.017mmHgSolution:- (C) 0.017mmHgLowering of vapour pressure =P0−p=p0.xsolute∴ΔP=35×0.6/600.660+36018=35×.01.01+20=35×.0120.01=0.17 mmHg

At room temperature, a dilute solution of urea is prepared by dissolving $0.60 g$ of urea in 360g of water. If the vapour pressure of pure water at this temperature is $35 m m H g$ . lowering of vapour pressure will be: (molar mass of urea $= 60 g m o l^{- 1}$ )

The correct option is B $0.017 m m H g$
Solution:- (C) $0.017 m m H g$

Lowering of vapour pressure $= P^{0} - p = p^{0} . x_{s o l u t e}$

$∴ Δ P = 35 \times \frac{0.6 / 60}{\frac{0.6}{60} + \frac{360}{18}}$

$= 35 \times \frac{.01}{.01 + 20} = 35 \times \frac{.01}{20.01}$

$= 0.17 m m H g$