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 = ${\displaystyle {\text{P}}_1^0}$ = 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: ${\displaystyle \frac{{\text{P}}_1^0-{\text{P}}_1}{{\text{P}}_1^0}=\frac{{\text{W}}_2{\text{M}}_1}{{\text{M}}_2{\text{W}}_1}}$

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,

${\displaystyle \frac{{\text{P}}_1^0-{\text{P}}_1}{{\text{P}}_1^0}=\frac{{\text{W}}_2{\text{M}}_1}{{\text{M}}_2{\text{W}}_1}}$

${\displaystyle =\frac{17{\text{mm Hg P}}_1}{17\text{mm Hg}}=\frac{2.8\text{g}\times 18{\text{g mol}}^{-1}}{50\text{g}\times 60{\text{g mol}}^{-1}}}$

∴ ${\displaystyle \frac{17{\text{mm Hg P}}_1}{17\text{mm Hg}}=0.0168}$

∴ 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.