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

Explanation

Here `P _(1) ^(0) =23.8mm`
`W _(2)= 50 g`
`M_(2) `(urea) `=60g mol^(-1)`
`W_(1)= 850g`
`M _(1) `(Water) `= 18g mol ^(-1)`
Here we have to calculate `P _(S)`
Applying Raoult.s law, `(P ^(0) - P_(S))/(P^(0))= (n_(2))/( n _(1) + n _(2)) = ((W _(2))/(M _(2)))/((W _(1))/(M _(1)) + (W_(2))/(M _(2)))`
` = ((50)/(60))/((850)/(18) + (50)/(60))= ( 0.83)/(48.05) =0.017`
Thus, relative lowering of vapour pressure `=0.017`
Substituting `P^(0) =23.8` mm Hg
`(23.8 -P_(s))/(P _(S)) = 0.017`
We get,
`23.8 - P_(S) = 0.017 P _(S)`
`P _(S) =23.4 mm Hg`
Thus, vapour pressure of water in the solution `=23.4` mm Hg