one gram of activated carbon has a surface are of `1000m^(2)`. Considering complete coverage as well as monomolecular adsorption , how much ammonia at 1 atm and 273 K would be absorbed on the surface of `44/7` g carbon if radius of a ammonia molecules is `10^(-8) cm`.

Options

A. `7.46 L`
B. `0.33 L `
C. `44. 8 L `
D. `23.5 L`

Answer

To solve the problem step by step, we will follow the given information and perform the necessary calculations.

### Step 1: Calculate the surface area of 44/7 grams of activated carbon.
Given that 1 gram of activated carbon has a surface area of 1000 m², we can find the surface area for 44/7 grams.

\[ \text{Surface Area} = \left(\frac{44}{7} \text{ g}\right) \times 1000 \text{ m}^2/\text{g} \]

Calculating this:

\[ \text{Surface Area} = \frac{44 \times 1000}{7} \text{ m}^2 = \frac{44000}{7} \text{ m}^2 \approx 6285.71 \text{ m}^2 \]

### Step 2: Calculate the surface area of a single ammonia molecule.
The surface area \( A \) of a sphere (ammonia molecule) is given by the formula:

\[ A = 4\pi r^2 \]

Where \( r \) is the radius of the ammonia molecule. Given \( r = 10^{-8} \) cm, we first convert it to meters:

\[ r = 10^{-8} \text{ cm} = 10^{-10} \text{ m} \]

Now substituting into the formula:

\[ A = 4 \pi (10^{-10})^2 = 4 \pi (10^{-20}) \approx 4 \times 3.14 \times 10^{-20} \approx 1.256 \times 10^{-19} \text{ m}^2 \]

### Step 3: Calculate the number of ammonia molecules that can be adsorbed.
To find the number of ammonia molecules that can be adsorbed, we divide the total surface area of the carbon by the surface area of one ammonia molecule:

\[ \text{Number of molecules} = \frac{\text{Total Surface Area}}{\text{Surface Area of one molecule}} = \frac{6285.71 \text{ m}^2}{1.256 \times 10^{-19} \text{ m}^2} \]

Calculating this gives:

\[ \text{Number of molecules} \approx \frac{6285.71}{1.256 \times 10^{-19}} \approx 5 \times 10^{22} \text{ molecules} \]

### Step 4: Calculate the volume of ammonia gas at STP.
At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters. We can convert the number of molecules to moles using Avogadro's number \( (6 \times 10^{23} \text{ molecules/mole}) \):

\[ \text{Moles of ammonia} = \frac{5 \times 10^{22}}{6 \times 10^{23}} \approx 0.0833 \text{ moles} \]

Now, we can find the volume of ammonia gas:

\[ \text{Volume} = \text{Moles} \times 22.4 \text{ L/mole} = 0.0833 \times 22.4 \approx 1.87 \text{ L} \]

### Final Answer:
The volume of ammonia that would be absorbed on the surface of \( \frac{44}{7} \) g of carbon is approximately **1.87 liters**.

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