Question

A catalyst has a total surface area of ${10}^3$ cm${}^2$ and adsorption sites available on it are $6\times$ 10${}^{15}$ sites per cm${}^2$. $H_2$ gas is adsorbed on this surface occupying only $5\%$ of the available sites of total surface area. On heating, the catalyst releases $H_2$ back and whole of the $H_2$ adsorbed is measured and found to occupy $2.46$ mL at $0.03$ atm and $300$ K. The number of $H_2$ molecules adsorbed on each site is:

• A. $6$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $6$

It is given that the catalyst on heating releases $H_2$ back and the whole of it is measured and found to occupy $2.46$ mL at $0.03$ atm and $300$ K.

Therefore, total number of moles of $H_2$ adsorbed $=\frac{PV}{RT}=\frac{0.03\times 2.46\times {10}^{-3}}{0.0821\times 300}=3\times {10}^{-6}$

$\therefore$ The number of $H_2$ molecules adsorbed $=3\times {10}^{-6}\times N_A=3\times {10}^{-6}\times 6\times {10}^{23}=18\times {10}^{17}$
Total number of sites available $=6\times {10}^{15}\times {10}^3=6\times {10}^{18}$

It is given that only $5\%$ of the available sites of total surface area are occupied.

$18\times {10}^{17}$ molecules were adsorbed on $\frac{6\times {10}^{18}\times 5}{100}$ sites $=3\times {10}^{17}$ sites
$\therefore$ The number of $H_2$ molecules/site $=\frac{18\times {10}^{17}}{3\times {10}^{17}}=6$

Hence, option A is the correct answer.