What weight of the non-volatile solute urea' `(NH_(2)-CO-NH_(2))` needs to be dissolved in `100 g` of water in order to decrease the vapour pressure of water by `25%`? What will be the molality of the solution?

Answer

`m=18.5`

Explanation

Let the vapour pressure `(VP)` of water=100
`:. `vapour pressure of urea solution=75
By Raoult's law
`(P^(@)-P_(S))/ P^(@)=(W_(2)//Mw_(2))/(W_(1)/(Mw_(1))+W_(2)/(Mw_(2))`
In Eq. (i), `W_(2)`(weight of urea)=?
`Mw_(2)`(molecular weight of urea)=60
`W_(1)`(weight of water)=100 g
`Mw_(2)`(molecular weight of water)=18 g
`P^(@)`(VP of water)=100
`P_(S)`(VP of solution)=75
So, `(100-75)/(100)=(W_(2)//60/(100/18+W_(2)/(60))`
On solving , we get ,`W_(2)=111.1 g`
Molality of urea solution =`"Moles of urea in gram"/"Weight of" `H_(2)O)` "in kg"`
=`(111.1//60)/(100//1000)=18.5M`