Volume of `N_(2)` at 1 atm, 273 K required to form a monolayer on the surface of iron catalyst is `8.15ml//gm` of the adsorbent. What will be the surface area of the adsorbent per gram if each nitrogen molecule occupies `16 xx 10^(-22)m^(2)`?
[Take : `N_(A)=6xx10^(23)`]

Options

A. `16 xx 10^(-16)cm^(2)`
B. `0.35 m^(2)//g`
C. `39 m^(2)//g`
D. `22400 cm^(2)`

Answer

To find the surface area of the adsorbent per gram, we will follow these steps:

### Step 1: Convert the volume of N₂ from milliliters to liters
Given that the volume of N₂ required to form a monolayer on the surface of the iron catalyst is 8.15 ml/g, we need to convert this volume into liters for our calculations.

\[ \text{Volume in liters} = 8.15 \, \text{ml} \times \frac{1 \, \text{L}}{1000 \, \text{ml}} = 0.00815 \, \text{L} \]

### Step 2: Use the Ideal Gas Law to find the number of moles of N₂
The Ideal Gas Law is given by the equation \( PV = nRT \), where:
- \( P \) = pressure (1 atm)
- \( V \) = volume (0.00815 L)
- \( n \) = number of moles
- \( R \) = gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (273 K)

Rearranging the equation to solve for \( n \):

\[ n = \frac{PV}{RT} \]

Substituting the values:

\[ n = \frac{(1 \, \text{atm}) \times (0.00815 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (273 \, \text{K})} \]

Calculating \( n \):

\[ n = \frac{0.00815}{22.414} \approx 3.63 \times 10^{-7} \, \text{mol} \]

### Step 3: Calculate the number of molecules of N₂
Using Avogadro's number (\( N_A = 6 \times 10^{23} \, \text{molecules/mol} \)), we can find the number of molecules:

\[ \text{Number of molecules} = n \times N_A = (3.63 \times 10^{-7} \, \text{mol}) \times (6 \times 10^{23} \, \text{molecules/mol}) \]

Calculating the number of molecules:

\[ \text{Number of molecules} \approx 2.18 \times 10^{17} \, \text{molecules} \]

### Step 4: Calculate the total surface area occupied by N₂ molecules
Given that each nitrogen molecule occupies \( 16 \times 10^{-22} \, \text{m}^2 \), we can find the total surface area:

\[ \text{Total surface area} = \text{Number of molecules} \times \text{Area per molecule} \]

Substituting the values:

\[ \text{Total surface area} = (2.18 \times 10^{17} \, \text{molecules}) \times (16 \times 10^{-22} \, \text{m}^2) \]

Calculating the total surface area:

\[ \text{Total surface area} \approx 0.35 \, \text{m}^2 \]

### Step 5: Surface area per gram of adsorbent
Since the calculations were done per gram of adsorbent, the surface area of the adsorbent per gram is:

\[ \text{Surface area per gram} = 0.35 \, \text{m}^2/\text{g} \]

### Final Answer
The surface area of the adsorbent per gram is approximately **0.35 m²/g**.

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