**Here, **

P°_{1} = 23.8 mm

W_{2} = 50 g

M_{2} (urea) = 60 g mol^{-1}

W_{1} = 850g

M_{1} (Water) = 18g mol^{-1}

Here we have to calculate P_{s}

**Applying Raoult’s law,**

**Thus,** relative lowering of vapour pressure = 0.017

Substituting P° = 23.8 mm Hg

\(\frac{23.8-P_S}{P_S}=0.017\)

**We get, **

23.8 – P_{s} = 0.017 P_{s}

P_{s} = 23.4 mm Hg

**Thus,** vapour pressure of water in the solution = 23.4 mm Hg