If 0.5 g of a mixture of two metals A and B with respective equivalent weights 12 and 9 displace 560 mL of H2 at STP from an acid, the composition of the mixture is

1 mol of H2=22400 mL=2Eq of H2 1Eq of H2 = 11200 mL Eq of H2 = 56011200=120Eq [Let the weight of A be x g; weight of B= 0.5 - x] Eq of A + Eq of B = Eq of H2 x12+0.5-x9=120 ∴x = 0.2 % of A = 0.2 ×1000.5=40% % of B = 60%