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 mm Hg, lowering of vapour pressure will be ______. (molar mass of urea = 60 g mol⁻¹)

पर्याय

• 0.031 mm Hg

• 0.017 mm Hg

• 0.028 mm Hg

• 0.027 mm Hg

Answer 1

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 Hg, lowering of vapour pressure will be 0.017 mm Hg.

Explanation:

Given: Mass of urea = 0.60 g

Molar mass of urea = 60 g/mol

Mass of water = 360 g = 0.360 kg

Vapour pressure of pure water, P° = 35 mm Hg

We are to find the lowering of vapour pressure

ΔP = P° − P 

ΔP = `P^circ * x_"solute"`   ...(i)

`"Moles of urea" (n_"urea") = 0.60/60` = 0.01 mol

`"Moles of water" (n_"water") = 360/18` = 20 mol

Mole fraction of urea = `0.01/(0.01 + 20)`

= `0.01/20.01`

= 0.0005

ΔP = 35 × 0.0005    ...[From equation (i)]

= 0.0175 mm Hg