Urea forms an ideal solution. Determine the vapour pressure of an aqueous solution containing `10` per cent by mass of urea at `40^(@)C=55.3mm Hg`).

Explanation

Given `10` per cent by mass of urea solution means `10g` urea is dissolved in `100g` of solution
Mass of solute (urea) `=10g`
Mass of solvent (water) `=90g`
Vapour pressure of pure solvent `55.3mm Hg`
Asked vapour pressure of solution `=?`
Formulae: `(P_(A)^(0)-P_(A))/(P_(A)^(0))=(W_(B)//M_(B))/(W_(A)/(M_(A))+(W_(B))/(M_(B)))`
Explanation: `P_(A)^(0)=` vapour pressure of pure solvent, `P_(A)=` vapour pressure of solution
`W_(B)=` Mass of solute, `M_(B)=` Molecular weight solute, `W_(A)=` Mass of solvent
`M_(A)=` Molecular weight of solvent
`(P_(A)^(0))/(P_(A)^(0))=(10//60)/(90/(18)+(10)/(60))=(1)/(31)`
`1-(P_(A))/(P_(A)^(0))=(1)/(31) rArr (P_(A))/(P_(A)^(0))=1-(1)/(31)=(30)/(31)`
`P_(A)=(55.3xx30)/(31)=(1659)/(31)=53.21mm Hg`.