Here,
P°1 = 23.8 mm
W2 = 50 g
M2 (urea) = 60 g mol-1
W1 = 850g
M1 (Water) = 18g mol-1
Here we have to calculate Ps
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 – Ps = 0.017 Ps
Ps = 23.4 mm Hg
Thus, vapour pressure of water in the solution = 23.4 mm Hg