Explanation

(b) $\frac{{m_{H}}_{2}}{M_{Cu}} = \frac{{W_{H}}_{2}}{W_{Cu}} = \frac{1}{\left( \ \frac{63.5}{2} \right)}$
$\Rightarrow {m_{H}}_{2} = \frac{2}{63.5} \times 0.3175 = 0.01$g
2g of $H_{2}$ occupies volume at NTP = 22.4 L
$\Rightarrow 0.01\ g\ of\ H_{2}$ occupies volume at NTP given by:
$\ V = \frac{22.4}{2} \times 0.01$L
$\Rightarrow V = 11.2 \times 0.01 \times 1000\ cc = 112\ cc$