Question

A mixture of aluminium and zinc weighing $1.67$ g was completely dissolved in acid and evolved $1.69$ litres of $H_2$ at NTP. The weight of aluminium in the original mixture is:

• A. $2.24$ g
• B. $1.24$ g
• C. $6.8$ g
• D. $4.2$ g

Answer 1

Answer: (B)

$1.24$ g

Aluminium and Zinc both evolve $H_2$ with acid.

Let $x$ g be the weight of Al in the mixture.

Then, weight of Zn $=(1.67-x)$ g.

Sum of equivalents of metals will be equal to the equivalents of hydrogen gas.
$(\frac{x}{Eq.wt.ofAl}+\frac{(1.67-x)}{Eq.wt.ofZn})=eq.$ of $H_2$
or, $\frac{x}{9}+\frac{1.67-x}{32.7}=\frac{1.69}{11.2}$ $($Eq.wt.of $Al=\frac{27}{9}$ and $Zn=\frac{65.4}{2},for{H}_2=\frac{22.4}{2}=11.2)$
$\therefore$ $x=1.24g$