How mich urea `("molar mass"=60 g mol^(-1))` must be dissolved in 50 g of water so that the vapour pressure at room temperature is reduced by 25% ? Also calculate the molality of the solution obtained.

Answer

`55.556g, 18.52m`

Explanation

If `p^(@)=100mm, p_(s)=75mm`. Hence, `=(100-75)/(100)=(w_(2)//60)/(50//18+w_(2)//60).` This gives `w_(2)=55.556g`
`"Molality of the solution "=(55.556//"60 mol")/(50 g)xx1000g kg^(-1)=18.52m`