Chief ore of `Zn` is `ZnS`. The ore is concentrated by froth flotation process and then heated in air to convert `ZnS` to `ZnO`. ltntgt `2 ZnS + 3O_(2) rarr 2 ZnO + 2 SO_(2)`…(i)
`ZnO`, thus formed is treated with dilute `H_(2) SO_(4)`.
`ZnO + H_(2) SO_(4) rarr ZnSO_(4) + H_(2) O` ....(ii)
On electrolysis of `ZnSO_(4(aq)), Zn` metal is produced.
`2 ZnSO_(4) + H_(2) O rarr 2 Zn + 2H_(2) SO_(4) + O_(2)`....(iii)
What volume of `98 % H_(2) SO_(4)` (by weight, density `= 1.8 g//mL`) is required in step `(ii)` ?

Options

A. 120 L
B. 129 L
C. 1.40 L
D. 150 L

Answer

B

Explanation

Molarity of `98% H_(2) So_(4)` by weight `H_(2) SO_(4)` whose `rho = 1.8 g//mL` can be calculated as follows :
`1 mL H_(2) SO_(4)` contains `= 1.8 g H_(2) SO_(4)`
`1000 ml H_(2) SO_(4)` contains `= 1.8 xx 10^(3) g H_(2) SO_(4)`
=`(1.8 xx 10^(3))/(98) kg ZnO`
i.e Molarity of `100 % H_(2) SO_(4) = 18.36 M`
Hence molarity of `98% H_(2) SO_(4) = (18.36)/(98) xx 98 = 18 M`
In Step (ii), `2.3195 kmol` of `ZnO` is present.
`1 mol` of `ZnO` reacts with `1 mol` of `H_(2) SO_(4)`.
Hence, `2.3195 kmol` of `ZnO` needs
`2.3195 kmol` of `H_(2) SO_(4)`
`18 mol` of `H_(2) SO_(4)` are present in `1 L`.
`2.3195 kmol` of `H_(2) SO_(4)` are present in
`(1)/(18) xx\2.3295 xx 10^(3) = 128.86 L~~ 129 L`.