The pressure of the gas was found to decrease from 720 to `480 mm Hg`. When `5g` of sample of activated charcoal was kept in a flask of one litre capacity maintained at `27^(@)C` . If the density of charcoal is `1.25 gm//mL`. The volume of gas adsorbed per `gm` of charcoal at `480 mm` of `Hg` is

Options

A. `80.03 mL`
B. `32.20 mL`
C. `100.61mL`
D. None of these

Explanation

To solve the problem, we need to determine the volume of gas adsorbed per gram of activated charcoal at a pressure of 480 mm Hg. We will follow these steps:

### Step 1: Convert the initial and final pressures from mm Hg to atm
- Initial pressure (P_initial) = 720 mm Hg
- Final pressure (P_final) = 480 mm Hg

To convert mm Hg to atm, we use the conversion factor \(1 \text{ atm} = 760 \text{ mm Hg}\).

\[ P_{\text{initial}} = \frac{720 \text{ mm Hg}}{760 \text{ mm Hg/atm}} \approx 0.9474 \text{ atm} \]

\[ P_{\text{final}} = \frac{480 \text{ mm Hg}}{760 \text{ mm Hg/atm}} \approx 0.6316 \text{ atm} \]

### Step 2: Calculate the number of moles of gas before and after adsorption using the ideal gas equation
The ideal gas equation is given by:

\[ PV = nRT \]

Where:
- \(P\) = pressure in atm
- \(V\) = volume in liters (1 L = 1 L)
- \(n\) = number of moles
- \(R\) = ideal gas constant = 0.0821 L·atm/(K·mol)
- \(T\) = temperature in Kelvin = 27°C + 273 = 300 K

#### Calculate initial moles (n_initial):
\[ n_{\text{initial}} = \frac{P_{\text{initial}} \times V}{R \times T} = \frac{0.9474 \text{ atm} \times 1 \text{ L}}{0.0821 \text{ L·atm/(K·mol)} \times 300 \text{ K}} \approx 0.0385 \text{ moles} \]

#### Calculate final moles (n_final):
\[ n_{\text{final}} = \frac{P_{\text{final}} \times V}{R \times T} = \frac{0.6316 \text{ atm} \times 1 \text{ L}}{0.0821 \text{ L·atm/(K·mol)} \times 300 \text{ K}} \approx 0.0256 \text{ moles} \]

### Step 3: Calculate the number of moles adsorbed
\[ n_{\text{adsorbed}} = n_{\text{initial}} - n_{\text{final}} = 0.0385 - 0.0256 = 0.0129 \text{ moles} \]

### Step 4: Calculate the volume of gas adsorbed at final pressure
Using the ideal gas equation again, we can find the volume of gas adsorbed:

\[ V_{\text{adsorbed}} = \frac{n_{\text{adsorbed}} \times R \times T}{P_{\text{final}}} \]

Substituting the values:

\[ V_{\text{adsorbed}} = \frac{0.0129 \text{ moles} \times 0.0821 \text{ L·atm/(K·mol)} \times 300 \text{ K}}{0.6316 \text{ atm}} \approx 0.50306 \text{ L} \]

### Step 5: Calculate the volume of gas adsorbed per gram of charcoal
Given that the sample mass of charcoal is 5 g:

\[ \text{Volume per gram} = \frac{V_{\text{adsorbed}}}{\text{mass of charcoal}} = \frac{0.50306 \text{ L}}{5 \text{ g}} \approx 0.10061 \text{ L/g} \]

To convert this to mL:

\[ 0.10061 \text{ L/g} = 100.61 \text{ mL/g} \]

### Final Answer
The volume of gas adsorbed per gram of charcoal at 480 mm Hg is approximately **100.61 mL/g**.

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