As we know that vapour pressure of water, $P_{1}^{\infty}=23.8 \mathrm{~mm}$ of $\mathrm{Hg}$

Weight of water, $w_{1}=850 \mathrm{~g}$

Weight of urea, $w_{2}=50 \mathrm{~g}$

Molecular weight of water, $M_{1}=18 \mathrm{~g} \mathrm{~mol}^{-1}$

Molecular weight of urea, $M_{2}=60 \mathrm{~g} \mathrm{~mol}^{-1}$

We must now determine the vapour pressure of water in the solution. Vapour pressure is denoted by.

As a result of Raoult’s law, we now have:

$\frac{P_{1}^{\circ}-P_{1}}{P_{1}^{\circ}}=\frac{n_{2}}{n_{1}+n_{2}}$

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

$\Rightarrow \frac{23.8-{{P}_{1}}}{23.8}=0.0173$

$\Rightarrow P_{1}=23.4 \mathrm{~mm}$ of $\mathrm{Hg}$ As a result, the vapour pressure of water in the given solution is $23.4 \mathrm{~mm}$ of $\mathrm{Hg}$ and its relative lowering is 0.0173.