Question

Silver (atomic weight = 108 g/mol) has a density of $5/\pi gc{m}^{-3}$. The number of silver atoms on a surface of area $1.35\times {10}^{-14}{m}^2$ can be expressed in scientific notation as $y\times {10}^x$. The value of $(x+y)$ is
(Take $N_A=6.0\times {10}^{23}$ and the value of $\pi =3$)

Answer 1

Let 'r' be the radius of the atom of silver
Volume of silver atom $=\frac{4}{3}\pi r^3$
$=\frac{Atomicmassofsilver}{Density\times N_A}$
$=\frac{108\times \pi }{5\times 6.0\times {10}^{23}}=3.6\pi \times {10}^{-23}c{m}^3$
Radius of silver atom, $r=(\frac{3.6\times \pi \times {10}^{-23}\times 3}{4\pi }{)}^{1/3}$
$=3.0\times {10}^{-8}cm$
circular area of silver atom, $\pi r^2$
$=\pi (3.0\times {10}^{-8}cm{)}^2$
$=27\times {10}^{-20}{m}^2$
Number of silver atoms in the surface $=\frac{Areaofthesurface}{Circularareaofasilveratom}$
$y\times {10}^x=\frac{1.35\times {10}^{-14}}{27\times {10}^{-20}}=5\times {10}^4$
$y=5$
$x=4$
$x+y=9$