Some Basic Concepts of Chemistry - Result Question 43

Answer:

Correct Answer: 43. (2)

Solution:

43. 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

$=\frac{\text { Number of sites }}{\text { Number of molecules }}=\frac{2 \times 6.023 \times 10^{16}}{6.023 \times 10^{16}}=2$