Adsorption is the presence of excess concentration of any particular component at the surface of liquid a solid phase as compared to bulk.This is due to presence of residual forces at the surface of body.In the adsorption of hydrogen gas over a sample of charcoal, `1.12cm^3` of `H_2(g)` measured over S.T.P was found to absorb per gram of charcoal.Consider only monolayer adsorption.Density of `H_2` is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100 ml of 0.5M `CH_3COOH` to form monolayer and thereby the molarity of `CH_3COOH` reduces to 0.49.
`((3xx47.43)/(4pi))^(1//3)=2.24`
`(2.24)^2=5`
Surface area of charcoal adsorbed by each molecule of (acetic soda) is-

Options

A. `7.8xx10^(-19) cm^2`
B. `7.8xx10^(-18) cm^2`
C. `7.8xx10^(-17) cm^2`
D. `7.8xx10^(-16) cm^2`

Explanation

To solve the problem of finding the surface area of charcoal adsorbed by each molecule of acetic acid, we will follow these steps:

### Step 1: Calculate the number of moles of acetic acid adsorbed
Given:
- Initial molarity of acetic acid = 0.5 M
- Final molarity of acetic acid = 0.49 M
- Volume of solution = 100 mL = 0.1 L

The change in molarity (ΔM) is:
\[ \Delta M = 0.5 - 0.49 = 0.01 \text{ M} \]

Now, calculate the moles of acetic acid adsorbed:
\[ \text{Moles adsorbed} = \Delta M \times \text{Volume} = 0.01 \text{ mol/L} \times 0.1 \text{ L} = 0.001 \text{ mol} \]

### Step 2: Calculate the number of molecules of acetic acid adsorbed
Using Avogadro's number (\(N_A = 6.022 \times 10^{23}\) molecules/mol):
\[ \text{Number of molecules} (n) = \text{Moles adsorbed} \times N_A = 0.001 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} = 6.022 \times 10^{20} \text{ molecules} \]

### Step 3: Calculate the specific surface area of charcoal
From the problem, we know that 1.12 cm³ of \(H_2(g)\) is adsorbed per gram of charcoal. We need to find the number of hydrogen molecules adsorbed.

Given:
- Volume of \(H_2\) = 1.12 cm³
- Molar volume of gas at STP = 22.4 L = 22400 cm³

Calculate moles of \(H_2\):
\[ \text{Moles of } H_2 = \frac{1.12 \text{ cm}^3}{22400 \text{ cm}^3/\text{mol}} = 5 \times 10^{-5} \text{ mol} \]

Now, calculate the number of \(H_2\) molecules:
\[ n' = 5 \times 10^{-5} \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} = 3.01 \times 10^{19} \text{ molecules} \]

### Step 4: Calculate the specific surface area of charcoal
Assuming the radius of hydrogen molecule (\(r\)) is calculated using the provided formula:
\[ r = \left(\frac{3}{4\pi} \cdot \frac{2}{0.07 \cdot 6.022 \times 10^{23}}\right)^{1/3} \approx 2.24 \times 10^{-8} \text{ cm} \]

Now, calculate the area occupied by one hydrogen molecule:
\[ \text{Area} = \pi r^2 = \pi (2.24 \times 10^{-8})^2 \approx 1.58 \times 10^{-15} \text{ cm}^2 \]

Now, calculate the specific surface area (SSA) of charcoal:
\[ \text{SSA} = n' \times \text{Area} = 3.01 \times 10^{19} \text{ molecules} \times 1.58 \times 10^{-15} \text{ cm}^2 \approx 4.77 \times 10^{4} \text{ cm}^2/\text{g} \]

### Step 5: Calculate the surface area occupied by each molecule of acetic acid
Finally, the area occupied by each molecule of acetic acid is:
\[ \text{Area per acetic acid molecule} = \frac{\text{SSA}}{n} = \frac{4.77 \times 10^{4} \text{ cm}^2/\text{g}}{6.022 \times 10^{20}} \approx 7.91 \times 10^{-16} \text{ cm}^2 \]

### Final Answer
The surface area of charcoal adsorbed by each molecule of acetic acid is approximately \(7.91 \times 10^{-16} \text{ cm}^2\).

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