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`
Molecule of acetic acids adsorbed-

Options

A. `6.023xx10^20`
B. `6.023xx10^21`
C. `6.023xx10^22`
D. `6.023xx10^23`

Explanation

To solve the problem regarding the adsorption of acetic acid on charcoal, we will follow these steps:

### Step 1: Calculate the initial moles of CH₃COOH before adsorption
Given:
- Molarity (C₁) = 0.5 M
- Volume (V) = 100 mL = 0.1 L

Using the formula for moles:
\[ \text{Moles before adsorption} = C_1 \times V = 0.5 \, \text{mol/L} \times 0.1 \, \text{L} = 0.05 \, \text{moles} \]

### Step 2: Calculate the final moles of CH₃COOH after adsorption
Given:
- Final molarity (C₂) = 0.49 M

Using the same volume:
\[ \text{Moles after adsorption} = C_2 \times V = 0.49 \, \text{mol/L} \times 0.1 \, \text{L} = 0.049 \, \text{moles} \]

### Step 3: Calculate the moles of CH₃COOH adsorbed
\[ \text{Moles adsorbed} = \text{Moles before adsorption} - \text{Moles after adsorption} = 0.05 - 0.049 = 0.001 \, \text{moles} \]

### Step 4: Convert moles of CH₃COOH adsorbed to number of molecules
Using Avogadro's number (Nₐ = \(6.023 \times 10^{23}\) molecules/mol):
\[ \text{Number of molecules adsorbed} = \text{Moles adsorbed} \times N_a = 0.001 \, \text{moles} \times 6.023 \times 10^{23} \, \text{molecules/mol} \]
\[ = 6.023 \times 10^{20} \, \text{molecules} \]

### Conclusion
The number of molecules of acetic acid adsorbed on 1 gram of charcoal is \(6.023 \times 10^{20}\).

### Final Answer
The correct option is **Option A: \(6.023 \times 10^{-20}\)**.

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