#### Question

Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH_{2}CONH_{2}) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

#### Solution

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

Weight of water taken, *w*_{1} = 850 g

Weight of urea taken, *w*_{2} = 50 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` = `(50/60)/(850/18 + 50/60`

`=>(23.8-p_1)/23.8` = `0.83/(47.22+0.83)`

`=>(23.8-p_1)/23.8`= 0.0173

`=> p_1 = 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