Molality(m) of a sulphuric acid solution in which the mol fraction of water is `0.85` is : Multiply your answer with 10 and then put value of it

Answer

98

Explanation

Mole fraction of water in solution = 0.85
Mole fraction of `H_(2)SO_(4)` in solution `= 1 - 0.85 = 0.15`
If `n_(1)` is the number of moles of water and `n_(2)` is the number of moles of `H_(2)SO_(4)` in the solution , then .
Mole fraction of `H_(2)SO_(4) = (n_(2))/(n_(1) + n_(2))= 0.15`
Molality of `H_(2)SO_(4)` solution means the number of moles of `H_(2)SO_(4)` present in 1000g of `H_(2)O` . Thus , we have .
`w_(1) = 100 g or n_(1) = (1000)/(18) = 55.55, n_(2) = ? `
`(n_(2))/(55.55+ n_(2))`
`n_(2) = 0.15n_(2) + 8.3325 or n_(2) = 9.8`
`therefore ` Molality = 9.8 m .