Question

At 25 °C, the saturated vapour pressure of water is 24 mm Hg. Find the saturated vapour pressure of a 5% aqueous solution of urea at the same temperature. (Molar mass of urea = 60 g mol⁻¹)

Answer 1

Given: Temperature = 25°C

5% aqueous solution of urea, i.e., 5 g of urea in 100 g of solution.

Mass of urea = 5% of 100 g = 5 g

Formula: Moles of urea = `"mass"/"molar mass"`

= `5/60`

= 0.0833 mol

Mass of water = 100 g − 5 g

= 95 g

Moles of water = `"mass"/"molar mass"`

= `95/18`

= 5.2778 mol

Mole fraction of water = `"Moles of water"/"Moles of water + Moles of urea"`

= `5.2778/(5.2778 + 0.0833)`

= `5.2778/(5.3611)`

= 0.9844

By using Raoult’s law,

`P_"solution"^circ = P_"water" × chi_"water"`

= 24 mm Hg × 0.9844

= 23.634 mm Hg