0.05 cm thick coating of silver is deposited on a plate of 0.05 m 2 area. The number of silver atoms deposited on plate are × 10 23 .( At mass Ag .=108, d =7.9 g cm -3 )

Chemistry · Electrochemistry

JEE Main 2024 — 30 January, Shift 1 — Question 75

0.05 cm thick coating of silver is deposited on a plate of $0.05 \mathrm{~m}^{2}$ area.

The number of silver atoms deposited on plate are $\qquad$ $\times 10^{23} .($ At mass Ag $\left.=108, \mathrm{~d}=7.9 \mathrm{~g} \mathrm{~cm}^{-3}\right)$

Answer: 11

Numerical answer — enter this value.

Step-by-step solution

Volume of silver coating $=0.05 \times 0.05 \times 10000$

$=25 \mathrm{~cm}^{3}$

Mass of silver deposited $=25 \times 7.9 \mathrm{~g}$

Moles of silver atoms $=\frac{25 \times 7.9}{108}$

Number of silver atoms $=\frac{25 \times 7.9}{108} \times 6.023 \times 10^{23}$

$=11.01 \times 10^{23}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 30 January, Shift 1
Subject: Chemistry
Chapter: Electrochemistry
Topic: Faraday's Laws