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$0.005\;cm$ thick coating of silver is deposited on a plate of $0.5 \;m^2$ area. The number of silver atoms deposited on the plate are (atomic mass of silver $=108$ , and its density $=7.9g cm^{-3}$

$(a)\;1.1 \times 10^{10} \\ (b)\;1.1 \times 10^{15}\\(c)\;1.1 \times 10^{24} \\(d)\;1.1 \times 10^{30}$

1 Answer

Volume of $Ag$ deposited $= (0.5 \times 100 \times 100)cm^2 \times 0.005\;cm=25\;cm$
Mass of $Ag$ deposited =Volume $\times$ density
$\qquad=25 \times 7.9=197.5$
$108\;gm$ of $Ag$ contain $=\large\frac{6.023 \times 10^{23}}{108} \times$$ 197$
$\qquad= 10.9 \times 10^{23}=1.1 \times 10^{24}$
Hence c is the correct answer.
answered Apr 1, 2014 by meena.p
 

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