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 is

Options

A. `40% A, 60% B`
B. `60% A, 40% B`
C. `30% A, 70% B`
D. `70% A, 30% B`

Answer

A

Explanation

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%`