Vapour pressure of pure water at 298 K is 23.8 mm Hg.

Solution:

It is given that vapour pressure of water, $p_{1}^{0}=23.8 \mathrm{~mm}$ of $\mathrm{Hg}$

Weight of water taken, w1 = 850 g

Weight of urea taken, w2 = 50 g

Molecular weight of water, M1 = 18 g mol−1

Molecular weight of urea, M2 = 60 g mol−1

Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p1.

Now, from Raoult’s law, we have:

$\frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=\frac{n_{2}}{n_{1}+n_{2}}$

$\Rightarrow \frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=\frac{\frac{w_{2}}{M_{2}}}{\frac{w_{1}}{M_{1}}+\frac{w_{2}}{M_{2}}}$

$\Rightarrow \frac{23.8-p_{1}}{23.8}=\frac{\frac{50}{60}}{\frac{850}{18}+\frac{50}{60}}$

$\Rightarrow \frac{23.8-p_{1}}{23.8}=\frac{0.83}{47.22+0.83}$

$\Rightarrow \frac{23.8-p_{1}}{23.8}=0.0173$

$\Rightarrow p_{1}=23.4 \mathrm{~mm}$ of $\mathrm{Hg}$

Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is 0.0173.