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

Explanation

Mole fraction of water = 0.85
Mole fraction of `H_(2)SO_(4)` in the solution `=1-0.85=0.15," i.e., "(n_(2))/(n_(1)+n_(2))=0.15`
where `n_(2)` is the number of mole of `H_(2)SO_(4)` and `n_(1)` is the number of moles of `H_(2)O` in the solution.
Molality of `H_(2)SO_(4)` solution means the number of moles of `H_(2)SO_(4)` present in 1000 g of `H_(2)O`. Thus, we have to find `n_(2)` when `w_(1)=1000`, i.e., `n_(1)=(1000)/(18)=55.55" moles."`
Substituting the value of `n_(1)` in equation (i), we get
`(n_(2))/(55.55+n_(2))=0.15" or "n_(2)=0.15n_(2)+8.3325 or 0.85n_(2)=8.3325`
`n_(2)=(8.3325)/(0.85)="9.8 moles, i.e, Molality"= 9.8m.`
Alternatively, `(n_(1))/(n_(1)+n_(2))=0.85" ...(i) "therefore" "(n_(2))/(n_(1)+n_(2))=1-0.85=0.15" ...(ii)"`
Dividing (ii) by (i), we get
`(n_(2))/(n_(1))=(0.15)/(0.85)or(n_(2))/(1000//18)=(0.15)/(0.85)orn_(2)=(0.15)/(0.85)xx(1000)/(18)=9.8" moles"`
`"Hence, molality = 9.8 m"`