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):

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): (A) 0.027 mmHg (B) 0.028 mmHg (C) 0.017 mmHg (D) 0.031 mmHg

Tardigrade Admin · Accepted Answer

Lowering of vapour pressure = p0 - p = p0 . xsolute ∴ Δ p =35 ×(0.6/60/(0.6)60 + (360)18 = 35 ×(0.1/.01+20) = 35×(.01/20.01) = .017 mm Hg

Q. 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 $35 mm H g$ , lowering of vapour pressure will be (molar mass of urea = $60 g$ mol $^{- 1}$ ):

0.027 mmHg

0.028 mmHg

0.017 mmHg

0.031 mmHg

Lowering of vapour pressure $= p^{0} - p = p^{0} . x_{so l u t e}$
$∴ Δ p = 35 \times \frac{0.6/60}{\frac{0.6}{60} + \frac{360}{18}}$
$= 35 \times \frac{0.1}{.01 + 20} = 35 \times \frac{.01}{20.01}$
= $.017 mm H g$

Q. 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 35mmHg, lowering of vapour pressure will be (molar mass of urea = 60g mol−1):