#### Question

30 g of urea (M = 60 g mol^{−1}) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23·8 mm Hg.

#### Solution

It is given that vapour pressure of water, `p_1^0` = 23.8 mm of Hg

Weight of water taken, _{w1} = 846 g

Weight of urea taken, w_{2} = 30 g

Molecular weight of water, M_{1} = 18 g mol^{−1}

Molecular weight of urea, M_{2} = 60 g mol^{−1}

Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p_{1}.

Now, from Raoult’s law, we have:

`(p_1^0 - p_1)/p_1^0 = n_2/(n_1 - n_2)`

`=> (p_1^0 - P_1)/p_1^0 = (w_2/M_2)/(w_1/M_1 - w_2/M_2)`

`= (23.8 - p_1)/23.8 = (30/60)/(346/18 + 30/60)`

`=> (23.8 - p_1)/23.8 = 0.0105`

=> p_{1} = 23.5501 mm of Hg

Hence, the vapour pressure of water in the given solution is 23.5501 mm of Hg and its relative lowering is 0.0105.