Question

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

Answer 1

$d =\frac{\text { mass }}{ V } $
$\Rightarrow 10.5\, g / cc$ means in $1 cc =10.5\, g$ of $Ag$ is present.
Number of atoms of $Ag$ in $1\, cc \Rightarrow \frac{10.5}{108} \times N _{ A }$
In $1 \,cm$, number of atoms of $Ag =\sqrt[3]{\frac{10.5}{108} N _{ A }}$
In $1 \,cm ^{2}$, number of atoms of $Ag =\left(\frac{10.5}{108} N _{A}\right)^{2 / 3}$
In $10^{-12} \,m ^{2}$ or $10^{-8}\, cm ^{2}$, number of atoms of
$Ag =\left(\frac{10.5}{108} N _{ A }\right)^{2 / 3} \times 10^{-8} $
$=\left(\frac{1.05 \times 6.022 \times 10^{24}}{108}\right)^{2 / 3} \times 10^{-8}$
$=1.5 \times 10^{7}$
Hence $x =7$