It is given that vapour pressure of water, p₁^o = 23.8 mm of 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,

\(\frac{p_1^0 - p_1}{p_1^0}\) = \(\frac{n_2}{n_1 + n_2}\)

⇒ \(\frac{p_1^0 - p_1}{p_1^0}\) = \(\frac{\frac{w_2}{M_2}}{\frac{w_1}{M_1} + \frac{w_2}{M_2}}\) 

⇒ \(\frac{23.8 - p_1}{23.8}\) = \(\frac{\frac{50}{60}}{\frac{850}{18} + \frac{50}{60}}\)

⇒ \(\frac{23.8 - p_1}{23.8}\) = \(\frac{0.83}{47.22 + 0.83}\)

⇒ \(\frac{23.8 - p_1}{23.8}\) = 0.0173

⇒ p₁ = 23.4 mm of Hg

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