Question

A sample of commercial sulphuric acid is `98% H_(2) SO_(4)` by mass. Calculate the mole fractions of `H_(2) SO_(4)` and `H_(2) O`.
A. `0.9, 0.1`
B. `0.1, 0.9`
C. `0.2, 0.8`
D. `0.8, 0.2`

Answer 1

Correct Answer - A
Consider `100g` of solution. Thus,
`m_(H_(2)) SO_(4) = 98g, m_(H_(2)O) = 2g`
`n_(H_(2)) SO_(4) = ("Mass of" H_(2) SO_(4))/("Molar mass of" H_(2) SO_(4)) = (98g)/(98g mol^(-1)) = 1 mol`
`n_(H_(2)O) = ("Mass of" H_(2) O)/("Molar mass of" H_(2)O) = (2g)/(18g mol^(-1)) = (1)/(9) mol`
`chi H_(2) SO_(4) = (.^(n)H_(2) SO_(4))/(.^(n)H_(2) SO_(4) + .^(n)H_(2)O) = (1mol)/((1 + 1//9) mol) = (1)/(10//9) = (9)/(10) = 0.9`
`chi H_(2) O = 1.0 - chi H_(2) SO_(4) = 1.0 - 0.9 = 0.1`