Question

Silver (atomic weight $=108gmo{l}^{-1}$ ) has a density of $10.5gc{m}^{-3}$. 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

Answer: 7

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