The vapour pressure of water at 20 °C is 17 mm Hg. What is the vapour pressure of solution containing 2.8 g urea in 50 g of water? - Chemistry

Exercise | Q 9 | Page 46

The vapour pressure of water at 20 °C is 17 mm Hg. What is the vapour pressure of solution containing 2.8 g urea in 50 g of water?

Solution:

Given: Vapour pressure of pure water = P10 = 17 mm Hg
Mass of urea (W2) = 2.8 g
Mass of water (W1) = 50 g

To find: Vapour pressure of the solution (P1)

Formula: P10-P1P10=W2M1M2W1

Calculation:

Molar mass of urea (NH2CONH2) = 14 + 2 + 12 + 16 + 14 + 2 = 60 g mol-1

Molar mass of water = 18 g mol–1

Now, using formula,

P10-P1P10=W2M1M2W1

=17mm Hg P117mm Hg=2.8g×18g mol-150g×60g mol-1

∴ 17mm Hg P117mm Hg=0.0168

∴ 17 mm Hg = 0.0168 × 17 mm Hg

∴ 17 mm Hg - P1 = 0.2856 mm Hg

∴ P1 = 17 mm Hg - 0.2856 mm Hg = 16.71 mm Hg

Vapour pressure of the given solution is 16.71 mm Hg.