Question

$20\%$ surface sites have absorbed $N_2$. On heating, $N_2$ gas is evolved from the sites and is collected at $0.001$ atm and $298$ K in a container of volume $2.46{cm}^3$. Density of surface sites is $6.023\times {10}^{14}{cm}^{-2}$ and surface area is $1000{cm}^2$. The number of surface sites occupied per molecule of $N_2$ is:

Answer 1

For absorbed $N_2$ on surface sites,
$P_{N_2}=0.001$ atm, $V=2.46$ cc $=2.46\times {10}^{-3}$ L, $T=298$ K

$n_{N_2}=\frac{PV}{RT}=\frac{0.001\times 2.46\times {10}^{-3}}{0.0821\times 298}=1.0\times {10}^{-7}$

$\therefore$ Molecules of $N_2$ absorbed $=1.0\times {10}^{-7}\times 6.023\times {10}^{23}$ $=6.023\times {10}^{16}$

$\therefore$ Total surface sites available $=$ Number of sites per ${cm}^2\times$ Area $=6.023\times {10}^{14}\times 1000=6.023\times {10}^{17}$

Surface sites on which $N_2$ is absorbed $=20\%\times$ Available sites
$=\frac{20}{100}\times 6.023\times {10}^{17}$ $=12.046\times {10}^{16}$

$\therefore$ Number of sites absorbed per molecule of $N_2$
$=\frac{12.046\times {10}^{16}}{6.023\times {10}^{16}}=2$