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If `0.5 g` of a mixture of two metals. `A` and `B` with respective equivalent weights 12 and 9 displace `560 mL` of `H_(2)` at `STP` from an acid, the composition of the mixture isA. `40% A, 60% B`B. `60% A, 40% B`C. `30% A, 70% B`D. `70% A, 30% B` |
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Answer» Correct Answer - A 1 mol of `H_(2) = 22400 mL = 2 Eq "of" H_(2)` `1 Eq of H_(2) = 11200 mL` Eq of `H_(2) = (560)/(11200) = (1)/(20) Eq` [Let of weight of `A` be weight of `B = 0.5 - x` Eq of `A + Eq "of" B = Eq "of" H_(2)` `(x)/(12) + (0.5)/(9) = (1)/(20)` % of `A = (0.2 xx 100)/(0.5) = 40%` % of `B = 60%` |
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