Some Basic Concepts of Chemistry - Result Question 43
43. $20 %$ surface sites have adsorbed $N _2$. On heating $N _2$ gas evolved from sites and were collected at $0.001 atm$ and 298 $K$ in a container of volume is $2.46 cm^{3}$. Density of surface sites is $6.023 \times 10^{14} / cm^{2}$ and surface area is $1000 cm^{2}$, find out the number of surface sites occupied per molecule of $N _2$.
(2005,3 M)
Show Answer
Answer:
Correct Answer: 43. (2)
Solution:
- Partial pressure of $N _2=0.001 atm$,
$T=298 K, V=2.46 dm^{3} .$
From ideal gas law : $p V=n R T$
$n\left(N _2\right)=\frac{p V}{R T}=\frac{0.001 \times 2.46}{0.082 \times 298}=10^{-7}$
$\Rightarrow$ Number of molecules of $N _2=6.023 \times 10^{23} \times 10^{-7}$
$=6.023 \times 10^{16}$
Now, total surface sites available
$=6.023 \times 10^{14} \times 1000=6.023 \times 10^{17}$
Surface sites used in adsorption $=\frac{20}{100} \times 6.023 \times 10^{17}$
$=2 \times 6.023 \times 10^{16}$
$\Rightarrow$ Sites occupied per molecules
$=\frac{\text { Number of sites }}{\text { Number of molecules }}=\frac{2 \times 6.023 \times 10^{16}}{6.023 \times 10^{16}}=2$
BETA
