$0.05 \mathrm{~cm}$ 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 ________ $\times 10^{23}$. (At mass $\mathrm{Ag}=108, \mathrm{~d}=7.9 \mathrm{~g} \mathrm{~cm}^{-3}$)
Answer (integer)
11
Solution
<p>$$\begin{aligned}
&\begin{aligned}
& \text { Volume of silver coating }=0.05 \times 0.05 \times 10000 \\\\
& =25 \mathrm{~cm}^3 \\\\
& \text { Mass of silver deposited }=25 \times 7.9 \mathrm{~g} \\\\
& \text { Moles of silver atoms }=\frac{25 \times 7.9}{108} \\\\
& \text { Number of silver atoms }=\frac{25 \times 7.9}{108} \times 6.023 \times 10^{23} \\\\
& =11.01 \times 10^{23}
\end{aligned}\\\\
&\text { Ans. } 11
\end{aligned}$$</p>
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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