Question

At 25°C, the vapour pressure of water is 23.75 mm Hg. Calculate the vapour pressure at the same temperature of 5% aqueous solution of urea (NH₂CONH₂). 

Answer 1

Given: Temperature = 25°C

Vapour pressure of pure water `P_"water"^circ` = 23.75 mm Hg

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

Molar mass of urea (NH₂CONH₂) = 60 g/mol

Solvent = Water (molar mass = 18 g/mol)

Since it’s a 5% w/w solution,

Mass of urea = 5 g

mass of water = 100 g – 5 g = 95 g

Moles of urea = `5/60` = 0.0833 mol

Moles of water = `95/18` = 5.278 mol

Mole fraction of water `chi_"water" = 5.278/(5.278 + 0.0833)`

= `5.278/5.3613`

= 0.9845

By using Raoult’s law,

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

= 0.9845 × 23.75

= 23.38 mm Hg

∴ The vapour pressure of 5% aqueous urea solution at 25°C is 23.38 mm Hg.