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

Explanation

Here,`" "p^(@)="23.8 mm"`
`" "w_(2)=50g," "M_(2)"(urea) = 60 g mol"^(-1)," "w_(1)=850 g, M_(1)" (water) = 18 g mol"^(-1)`
Our aim is to calculate `p_(s) and (p^(@)-p_(s))//p^(@)`
`"Applying Raoult's law."=(p^(@)p_(s))/(p^(@))=(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)/(47.22+0.83)=(0.83)/(48.05)=0.17`
Thus, relative lowering of vapour pressure = 0.017
Substituting `p^(@)=23.8` mm, we get
`(23.8-p_(s))/(p^(@))=0.017 or(23.8-p_(s))/(23.8)=0.017or23.8-p_(s)=0.017 xx23.8=0.40`
`"or "p_(s)=23.8-0.40="23.4 mm"`
Thus, vapour pressure of water in the solution = 23.4 mm.