At $25^{o}$ C the saturated vapour pressure of water is $3.165$ kPa ( $23.75$ mm Hg). Find the saturated vapour pressure of a $5 %$ aqueous solution of urea (carbamide) at the same temperature. (Molar mass of urea $= 60.05$ g $m o l^{- 1}$ )

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Solution

Let us consider the problem:
Given that:
$W = 5 g, W_{A} = 95 g, M_{B} = 60.05 g m o l^{- 1}$
$M_{A} = 18 g m o l^{- 1}, P^{0}_{A} = 3.165 k P a$
Putting the values in expression
$\frac{P_{A}^{o} - P}{P_{A}^{o}} = \frac{W_{B} \times M_{A}}{M_{B} \times W_{A}}$
$\frac{3.165 k P a - P}{3.165 k P a} = \frac{5 g \times 18 g m o l^{- 1}}{60.05 g m o l^{- 1} \times 95 g} = 0.015$
Hence,
$P = 3.165 k P a - 0.015 \times 3.165 k P a$
$P = 3.118 k P a$ .

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