It is given that vapour pressure of water, po = 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:

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