Calculate the surface area of a catalyst that adsorbs `10^3 cm^3` of `N_2` (reduced of STP) per gram in order to form the monolayer.The effective area occupied by `N_2` molecule of the surface is`1.62×10^(−15)cm^2`

Answer

To calculate the surface area of a catalyst that adsorbs \(10^3 \, \text{cm}^3\) of \(N_2\) per gram in order to form a monolayer, we can follow these steps:

### Step 1: Calculate the number of \(N_2\) molecules
To find the number of \(N_2\) molecules adsorbed, we can use the formula:

\[ \text{Number of molecules} = \frac{\text{Volume of } N_2 \, (\text{in cm}^3) \times N_A}{\text{Volume of 1 mole of gas at STP (in cm}^3\text{)}} \]

Where:
- \(N_A\) (Avogadro's number) = \(6.023 \times 10^{23} \, \text{molecules/mol}\)
- Volume of 1 mole of gas at STP = \(22400 \, \text{cm}^3\)

Substituting the values:

\[ \text{Number of molecules} = \frac{10^3 \, \text{cm}^3 \times 6.023 \times 10^{23} \, \text{molecules/mol}}{22400 \, \text{cm}^3} \]

Calculating this gives:

\[ \text{Number of molecules} \approx \frac{10^3 \times 6.023 \times 10^{23}}{22400} \approx 2.69 \times 10^{22} \, \text{molecules} \]

### Step 2: Calculate the total area covered by \(N_2\)
The total area covered by the \(N_2\) molecules can be calculated using the formula:

\[ \text{Total area} = \text{Number of molecules} \times \text{Area occupied by one molecule} \]

Where the area occupied by one \(N_2\) molecule is given as \(1.62 \times 10^{-15} \, \text{cm}^2\).

Substituting the values:

\[ \text{Total area} = 2.69 \times 10^{22} \, \text{molecules} \times 1.62 \times 10^{-15} \, \text{cm}^2 \]

Calculating this gives:

\[ \text{Total area} \approx 2.69 \times 10^{22} \times 1.62 \times 10^{-15} \approx 4.35 \times 10^{7} \, \text{cm}^2 \]

### Step 3: Convert the total area to square meters (if needed)
To convert the area from \(\text{cm}^2\) to \(\text{m}^2\):

\[ \text{Total area in } \text{m}^2 = \frac{4.35 \times 10^{7} \, \text{cm}^2}{10^4} = 4.35 \times 10^{3} \, \text{m}^2 \]

### Final Answer
The surface area of the catalyst that adsorbs \(10^3 \, \text{cm}^3\) of \(N_2\) per gram in order to form a monolayer is approximately \(4.35 \times 10^{7} \, \text{cm}^2\) or \(4.35 \times 10^{3} \, \text{m}^2\).

---