Mass in grams of copper deposited by passing $9.6487$ A current through a voltmeter containing copper sulphate solution for $100$ seconds is: (Given: Molar mass of $\mathrm{Cu}$ $=63$ g mol${}^{-1}$, $1F=96487$ C)

Solution & Explanation

\textbf{Answer:} (A) 0.315 g \textbf{Solution:} At the cathode: $\mathrm{Cu}{}^{2+}+2\mathrm{e}{}^{-}\to \mathrm{Cu}$ ($n=2$). Charge passed: $Q=It=9.6487\times 100=964.87\text{C}$. Using Faraday's first law $w=\frac{M\cdot Q}{n\cdot F}$, $$w=\frac{63\times 964.87}{2\times 96487}=\frac{964.87}{96487}\times \frac{63}{2}=0.01\times 31.5=0.315\text{g}.$$