Calculate the molarity of sulphuric acid solution in which mole fraction water is 0.85.

Answer

9.8

Explanation

Mole fraction of water = 0.85, Mole fraction of `H_(2)SO_(4)=1 -0.85 = 0.15`
Let `n_(B)` moles of `H_(2)SO_(4)` be dissolved in 1000 g of water to represent the molality of the solution
`:.` No. of moles of water `(n_(A))=((1000g))/(("18 g mol"^(-1)))=55.55` mol
No. of moles of `H_(2)SO_(4)=n_(B)`
`(n_(B))/(n_(A)+n_(B))=0.15or(n_(B))/(n_(B)+55.55)=0.15`
`n_(B)=0.15 n_(B)+55.5xx0.15or`
`n_(B)=(55.5xx0.15)/(0.85)=9.8`.