Chemistry · Electrochemistry

JEE Main 2024 — 29 January, Shift 1 — Question 74

The mass of zinc produced by the electrolysis of zinc sulphate solution with a steady current of 0.015 A for 15 minutes is $\qquad$ $\times 10^{-4} \mathrm{~g}$ . (Atomic mass of zinc $=65.4 \mathrm{amu})$

Answer: 45.75

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{Zn}^{+2}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}$

$\mathrm{W}=\mathrm{Z} \times \mathrm{i} \times \mathrm{t}$

$=\frac{65.4}{2 \times 96500} \times 0.015 \times 15 \times 60$

$=45 \cdot 75 \times 10^{-4} \mathrm{gm}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Chemistry
Chapter: Electrochemistry
Topic: Faraday's Laws