Silver (atomic weight $=108 \mathrm{~g} \mathrm{~mol}^{-1}$ ) has a density of $10.5 \mathrm{~g} \ma

Question

Silver (atomic weight $=108 \mathrm{~g} \mathrm{~mol}^{-1}$ ) has a density of $10.5 \mathrm{~g} \mathrm{~cm}^{-3}$. The number of silver atoms on a surface of area $10^{-12} \mathrm{~m}^{2}$ can be expressed in scientific notation as $y 	imes 10^{x}$. The value of $x$ is

Professor PC Thomas Classes · Answer

(7) : Given, atomic weight $ =108 \mathrm{~g} \mathrm{~mol}^{-1} $ Density $=10.5 \mathrm{~g} \mathrm{~cm}^{-3}$,\ Surface area $=10^{-12} \mathrm{~m}^{2}$\ Volume of one silver atom $=4 / 3 \pi r^{3}$\ $\because$ Density $=\frac{	ext { Mass }}{	ext { Volume }}$\ $\Rightarrow$ Volume $=\frac{	ext { Mass }}{	ext { Density }}$\ or $\frac{4}{3} \pi r^{3}=\frac{108}{6.023 	imes 10^{23} 	imes 10.5}$\ $r^{3}=\frac{108 	imes 3}{6.023 	imes 10^{23} 	imes 10.5 	imes 4 	imes 3.14}$\ $r^{3}=0.40 	imes 10^{-23}=4 	imes 10^{-24}$\ or $r=1.58 	imes 10^{-8} \mathrm{~cm}$ or $1.58 	imes 10^{-10} \mathrm{~m}$\ No. of silver atoms on a surface area of $10^{-12} \mathrm{~m}^{2}$ can be given by $10^{-12}=\pi r^{2} 	imes n$\ $n=\frac{10^{-12}}{3.14 	imes\left(1.58 	imes 10^{-10}ight)^{2}}=0.127 	imes 10^{8}$\ $\Rightarrow n=1.27 	imes 10^{7}$ or $x=7$\