Vapour pressure in mm Hg of 0.1 mole of urea in 180 g of water at 25° C is (The vapour pressure of water at 25° C is 24 mm Hg)

Question

Vapour pressure in mm Hg of 0.1 mole of urea in 180 g of water at 25° C is (The vapour pressure of water at 25° C is 24 mm Hg)  (A) 2.376 (B) 20.76 (C) 23.76 (D) 24.76

Tardigrade Admin · Accepted Answer

From Raoult's law for very dilute solution. (p°-ps/p°)=(w/m) × (M/W) (24-ps/24)=0.1 × (18/180) 24-ps=0.24 ∴ ps=24-0.24=23.76 mm Hg

Q. Vapour pressure in mm $H g$ of $0.1$ mole of urea in $180 g$ of water at $2 5^{\circ} C$ is (The vapour pressure of water at $2 5^{\circ} C$ is $24 mm H g)$

2.376

20.76

23.76

24.76

From Raoult's law for very dilute solution.

$\frac{p ^{\circ} - p _{s}}{p ^{\circ}} = \frac{w}{m} \times \frac{M}{W}$

$\frac{24 - p _{s}}{24} = 0.1 \times \frac{18}{180}$

$24 - p_{s} = 0.24$

$∴ p_{s} = 24 - 0.24 = 23.76 mm H g$

Q. Vapour pressure in mm Hg of 0.1 mole of urea in 180g of water at 25∘C is (The vapour pressure of water at 25∘C is 24mmHg)