Chemistry · Electrochemistry

JEE Main 2024 — 29 January, Shift 2 — Question 83

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 g . The total charge passed through the solution is $\qquad$ $\times 10^{-2} \mathrm{~F}$ . (Given atomic mass of $\mathrm{Au}=197$ )

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

$\frac{\mathrm{W}}{\mathrm{E}}=\frac{\text { ch arge }}{1 \mathrm{~F}}$

$\begin{matrix}\frac{1.314}{\frac{197}{3}}=\frac{\text{Q}}{1\text{ }\!\!~\!\!\text{ F}} \\\text{Q}=2\times {{10}^{-2}}\text{ }\!\!~\!\!\text{ F} \\\end{matrix}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 2
Subject: Chemistry
Chapter: Electrochemistry
Topic: Faraday's Laws