Given Data:

P = 0.01 atm

T = 298 K

V = 2.46 cm$$^3$$

Density of surface sites $$=6.023\times10^{14}/\mathrm{cm^3}$$

Surface area (A) = 1000 cm$$^2$$

Goal:

We want to find out how many surface sites are covered by each nitrogen (N$$_2$$) molecule.

Step 1: Find the Total Number of Surface Sites

We are given the number of sites per cm$$^3$$ and the total surface area. Multiply these to get the total number of sites:

Total number of surface sites = Density $\times$ Total surface area

$$=6.023\times10^{14}\mathrm{~cm}^{-3}\times1000\mathrm{~cm^2}$$

$$=6.023\times10^{17}$$ sites

Step 2: Calculate the Number of Sites Occupied by N$$_2$$

Nitrogen covers just 20% of the sites. So, multiply the total sites by 20% (or 0.20):

Sites occupied by N$$_2$$ = $$\frac{20}{100}\times6.023\times10^{17}$$

$$=1.2046\times10^{17}$$ sites

Step 3: Find Number of N$$_2$$ Molecules Released (Using Gas Laws)

To get the number of N$$_2$$ molecules, first calculate how many moles of N$$_2$$ gas are collected using the ideal gas law:

$$n = \frac{PV}{RT}$$

Where R = 0.0821 L-atm/K·mol; V = 2.46 cm$^3$ = 2.46 × 10$^{-3}$ L

Now, plug in the values:

$$n = \frac{0.01\,\text{atm} \times 2.46 \times 10^{-3}\,\text{L}}{0.0821\, \text{L-atm/(K·mol)} \times 298\,\text{K}}$$

$$n = 1.01 \times 10^{-7}$$ moles

Now, multiply the number of moles by Avogadro's number (6.023 × 10$^{23}$) to get molecules:

Total N$$_2$$ molecules released = $n \times N_A$

$$= 1.01\times 10^{-7}\times 6.023\times10^{23} = 6.08\times10^{16}$$ molecules

Step 4: Number of Surface Sites Occupied by Each N$$_2$$ Molecule

Now divide the number of sites occupied by the number of N$$_2$$ molecules:

$$\text{Number of sites per N}_2 = \frac{1.2046\times10^{17}}{6.08\times10^{16}} = 1.98 \approx 2$$

Final Answer

Each nitrogen molecule occupies about 2 surface sites.

JEE Advanced Subjects