An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent . What is the molecular mass of the solute ?

Answer

`Mw_(2)=41.35 g mol^(-1)`

Explanation

`p^(@)=1 atm=1.013 ba r, p_(S)=1.004 ba r`
`W_(2)=2g,W_(sol)=100g,W_(1)=100-2=98g`
For dilute solution , use the relation `:`
`(p^(@)-p_(S))/(P^(@))=(n_(2))/(n_(1)+n_(2))=(n_(2))/(n_(1))=(W_(2)//Mw_(2))/(W_(1)//Mw_(1))`
`((1.013-1.004)ba r)/(1.013ba r)=(2g)/(Mw_(2))xx(18g mol^(-1))/(98g)`
`Mw_(2)=(2xx18)/(98)xx(1.013)/(0.009)g mol^(-1)`
`=41.35g mol^(-1)`