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

# 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. A

2. B

3. C

4. D

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### Solution:

According to STP conditions, one mole of a gas has a volume of 22400 mL. At STP conditions, one mole of H2 has a volume of 22400 mL.

If one mole of H2 (or 2 equivalents of H2) has a volume of 22400 mL, then one equivalent of H2 has a volume of 11200 mL.

If one equivalent of H2 has a volume of 11200 mL, then the number of equivalents of H2 present in 560 mL is calculated as:

Total mass of the mixture is 5.6 g. Assume that the mass of A is x g and the mass of B is 0.5-x g.

Equivalent mass of A is calculated as:

Equivalent mass of B is calculated as:

Equivalents of A and B displaces the equivalents of H2. This implies that sum of equivalents of A and B is equal to the equivalents of H2 present in 560 mL.

Substitute the values in the above equation and determine the value of x.

$\begin{array}{ccc}\frac{\mathrm{x}}{12}+\frac{0.5-\mathrm{x}}{9}& =& \frac{1}{20}\\ \frac{3\mathrm{x}+4\left(0.5-\mathrm{x}\right)}{36}& =& \frac{1}{20}\\ \mathrm{x}& =& 0.2\end{array}$

Therefore, mass of A is 0.2 g and the mass of B is 0.3 g (0.5 g-0.2 g).

Composition of A in the mixture is calculated as:

Composition of B in the mixture is calculated as:

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