Question

When a quantity of electricity is passed through $CuSO_4$ solution, $0.16\, g$ of Copper gets deposited. If the same quantity of electricity is passed through acidulated water, then the volume of $H_2$ liberated at $STP$ will be [Given $At. Wt. Cu = 64$]

• A. $4.0 \,cm^3$
• B. $56\, cm^3$
• C. $604 \,cm^3$
• D. $8.0\, cm^3$

Answer 1

Answer: (B)

$\frac{\text { Wt. of Cu deposited }}{\text { Wt. of } H _{2} \text { produced }}=\frac{\text { Eq. } \text { wt. of } Cu }{\text { Eq. wt. of } H } $

$\frac{0.16}{ wt \text { . of } H _{2}}=\frac{64 / 2}{1}=\frac{32}{1} $

wt. of $ H _{2}=\frac{0.16}{32}=5 \times 10^{-3}\, g$

Volume of $H _{2}$ liberated at STP

$=\frac{22400}{2} \times 5 \times 10^{-3} \,cc =56\, cc$