JEE MAIN - Chemistry (2024 - 29th January Evening Shift - No. 25)
A constant current was passed through a solution of $$\mathrm{AuCl}_4^{-}$$ ion between gold electrodes. After a period of 10.0 minutes, the increase in mass of cathode was $$1.314 \mathrm{~g}$$. The total charge passed through the solution is _______ $$\times 10^{-2} \mathrm{~F}$$.
(Given atomic mass of $$\mathrm{Au}=197$$)
Answer
2
Explanation
$$\begin{aligned}
& \frac{\mathrm{W}}{\mathrm{E}}=\frac{\text { charge }}{1 \mathrm{~F}} \\
& \frac{1.314}{\frac{197}{3}}=\frac{\mathrm{Q}}{1 \mathrm{F}} \\
& \mathrm{Q}=2 \times 10^{-2} \mathrm{~F}
\end{aligned}$$
Comments (0)
