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

`P^(@)=23.8mm`
`w_(2)=50,M_(2)("urea")=60" g "mol^(-1)`
`w_(1)=850g,M_(1)("water")=18" g "mol^(-1)`
To find `P_(S) and (P^(@)-P_(S))//P^(@)`
Solution: Applying raoult's law
`(P^(@))/(P^(@))=(n_(2))/(n_(1)+n_(2))=(w_(2)//M_(2))/(w_(1)//M_(1)+w_(2)//M_(2))`
`therefore(P^(@)-P_(S))/(P^(@))=(50//60)/(850//18+50//60)`
`=(0.83)/(47.22+0.83)=0.017`
Putting `P^(@)=23.8mm`, we have
`(23.8-P_(S))/(P_(S))=0.017`
`implies23.8-P_(S)=0.017P_(S)`
or, `1.017P_(S)=23.8`
or, `P_(S)=23.4mm`