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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)` ?A. 120 LB. 129 LC. 1.40 LD. 150 L |
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Answer» Correct Answer - B 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) xx2.3295 xx 10^(3) = 128.86 L~~ 129 L`. |
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