If 0.5 g of a mixture of two metals A and B | vedclass

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(A) $1 \ mol$ of $H_2 = 22400 \ mL = 2 \ Eq$ of $H_2$. $1 \ Eq$ of $H_2 = 11200 \ mL$. $Eq$ of $H_2 = \frac{560}{11200} = \frac{1}{20} \ Eq$. Let the

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$40\% \ A, 60\% \ B$

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(A) $1 \ mol$ of $H_2 = 22400 \ mL = 2 \ Eq$ of $H_2$. $1 \ Eq$ of $H_2 = 11200 \ mL$. $Eq$ of $H_2 = \frac{560}{11200} = \frac{1}{20} \ Eq$. Let the weight of $A$ be $x \ g$,then the weight of $B = (0.5 - x) \ g$. Since $Eq$ of $A + Eq$ of $B = Eq$ of $H_2$: $\frac{x}{12} + \frac{0.5 - x}{9} = \frac{1}{20}$. Multiplying by $36$: $3x + 4(0.5 - x) = 36 	imes \frac{1}{20} = 1.8$. $3x + 2 - 4x = 1.8$. $-x = -0.2$,so $x = 0.2 \ g$. $\% \ A = \frac{0.2}{0.5} 	imes 100 = 40\%$. $\% \ B = 100\% - 40\% = 60\%$.

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:

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