Q29 / 180 · COMEDK 2018 · Chemistry
When a quantity of electricity is passed through $\mathrm{CuSO}_{4}$ solution, $0.16 \mathrm{~g}$ of copper gets deposite(d)If the same quantity of electricity is passed through acidulated water, then the volume of $\mathrm{H}_{2}$ liberated at STP will be [Given, atomic weight of $\mathrm{Cu}=64$ ]
- A$4.0 \mathrm{~cm}^{3}$
- Correct answerB$56 \mathrm{~cm}^{3}$
- C$604 \mathrm{~cm}^{3}$
- D$8.0 \mathrm{~cm}^{3}$
Solution
According to second law of Faraday,
$$
\begin{aligned}
\frac{w_{1}}{E_{1}} &=\frac{w_{2}}{E_{2}} \\
\frac{0.16}{64} &=\frac{w_{2}}{2} \\
w_{2} &=\frac{0.16 \times 2}{64}=\frac{0.01}{2}=0.005 \mathrm{~g}
\end{aligned}
$$
Volume of $\mathrm{H}_{2}=\frac{0.005}{2} \times 22.400 \mathrm{~cm}^{3}=\frac{22.4}{4}=56 \mathrm{~cm}^{3}$
##0
Hence, the correct answer is option: 'B'.