0.05 cm thick coating of silver is deposited | vedclass

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(D) Volume of silver coating = $0.05 \ cm 	imes (0.05 \ m^2 	imes 10^4 \ cm^2/m^2) = 0.05 \ cm 	imes 500 \ cm^2 = 25 \ cm^3$.Mass of silver deposit

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(D) Volume of silver coating = $0.05 \ cm 	imes (0.05 \ m^2 	imes 10^4 \ cm^2/m^2) = 0.05 \ cm 	imes 500 \ cm^2 = 25 \ cm^3$.Mass of silver deposited = $	ext{Volume} 	imes 	ext{Density} = 25 \ cm^3 	imes 7.9 \ g \ cm^{-3} = 197.5 \ g$.Moles of silver atoms = $\frac{	ext{Mass}}{	ext{Atomic mass}} = \frac{197.5}{108} \approx 1.8287 \ 	ext{mol}$.Number of silver atoms = $	ext{Moles} 	imes N_A = 1.8287 	imes 6.022 	imes 10^{23} \approx 11.01 	imes 10^{23}$.Thus,the number of silver atoms is approximately $11 	imes 10^{23}$.

$0.05 \ cm$ thick coating of silver is deposited on a plate of $0.05 \ m^2$ area. The number of silver atoms deposited on the plate is . . . . . . $\times 10^{23}$. (Atomic mass of $Ag = 108$,density $d = 7.9 \ g \ cm^{-3}$)

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