Question

Silver (Atomic weight = 108 $\text{g/mol}$) has a density of 5.5 ${\text{g/cm}}^3$. The number of silver atoms on a surface of area ${10}^{-12}{\text{m}}^2$ can be expressed in scientific notation as $\text{8.3}\times {10}^x$. The value of x is:
( Take: $N_A=6\times {10}^{23},\pi =3)$

Answer 1

Volume of one $Ag$ atom = $\frac{4}{3}\pi r^3$
$\frac{\text{mass}}{\text{density}}$ = $\frac{4}{3}\pi r^3$
$\therefore \frac{108}{6\times {10}^{23}\times 5.5}$ = $\frac{4}{3}\pi r^3$
$r^3=0.818\times {10}^{-23}\text{cm}$
$r\approx 2\times {10}^{-8}\text{cm}$
$r=2\times {10}^{-10}\text{m}$
$\therefore$ Number of $Ag$ atoms in ${10}^{-12}{\text{m}}^2$
$\pi r^2\times n={10}^{-12}{\text{m}}^2$
$n=\frac{{10}^{-12}}{3\times (2\times {10}^{-10}{)}^2}$
$n=\frac{{10}^8}{12}$
$n=0.083\times {10}^8$
$n=8.3\times {10}^6$
The value of $\text{x}=6$