Question

At room temperature, a dilute solution of urea is prepared by dissolving $40.60\, g$ of urea in $360\, g$ of water. If the vapour pressure of pure water at this temperature is $35\, mmHg$, what will be the lowering of vapour pressure? (molar mass of urea $= 60 \,g\, mol^{-1}$)

Answer 1

Relative lowering of vapour pressure, is given
by, $\frac{P^{o} -P}{P^{o}} = x_{A} = \frac{n_{A}}{n_{A} +n_{B}} \simeq \frac{n_{A}}{n_{B}}$
Given, $P^{o} = 35 mm Hg, n_{\text{urea}} = \frac{0.60}{60}$
$n_{\text{water}} = \frac{360}{18}$
$\frac{P^{o} -P}{35} = \frac{0.6 \times18}{60 \times360} = \frac{1}{2000}$
$\Delta P =P^{o} -P = 0.017$