A sample of charcoal weighing `6 g` was brought into contact with a gas contained in a vessel of one litre capacity at `27^(@)C`. The pressure of the gas was found to fall from `700` to `400` mm. Calculate the volume of the gas (reduced to STP) that is adsorbent under the condition of the experiment `("density of charcoal sample is" 1.5 g cm^(3))`.

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the initial and final volume of the gas
Given:
- Initial pressure (P1) = 700 mmHg
- Final pressure (P2) = 400 mmHg
- Volume of the vessel (V1) = 1 L = 1000 mL

Using the formula for gas laws at constant temperature:
\[ P_1 V_1 = P_2 V_2 \]

We can rearrange this to find \( V_2 \):
\[ V_2 = \frac{P_1 V_1}{P_2} \]

Substituting the values:
\[ V_2 = \frac{700 \, \text{mmHg} \times 1000 \, \text{mL}}{400 \, \text{mmHg}} \]
\[ V_2 = \frac{700000}{400} = 1750 \, \text{mL} \]

### Step 2: Calculate the volume of charcoal
Given:
- Mass of charcoal = 6 g
- Density of charcoal = 1.5 g/cm³

Using the formula for density:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
We can rearrange this to find the volume of charcoal:
\[ \text{Volume of charcoal} = \frac{\text{Mass}}{\text{Density}} \]

Substituting the values:
\[ \text{Volume of charcoal} = \frac{6 \, \text{g}}{1.5 \, \text{g/cm}^3} = 4 \, \text{cm}^3 = 4 \, \text{mL} \]

### Step 3: Calculate the actual volume available for gas
The actual volume available for the gas in the vessel after accounting for the volume occupied by the charcoal:
\[ \text{Actual volume} = \text{Total volume} - \text{Volume of charcoal} \]
\[ \text{Actual volume} = 1000 \, \text{mL} - 4 \, \text{mL} = 996 \, \text{mL} \]

### Step 4: Calculate the volume of gas adsorbed
The volume of gas adsorbed can be calculated as:
\[ \text{Volume of gas adsorbed} = V_2 - \text{Actual volume} \]
\[ \text{Volume of gas adsorbed} = 1750 \, \text{mL} - 996 \, \text{mL} = 754 \, \text{mL} \]

### Step 5: Calculate the volume of gas adsorbed per gram of charcoal
To find the volume of gas adsorbed per gram of charcoal:
\[ \text{Volume per gram} = \frac{\text{Volume of gas adsorbed}}{\text{Mass of charcoal}} \]
\[ \text{Volume per gram} = \frac{754 \, \text{mL}}{6 \, \text{g}} = 125.67 \, \text{mL/g} \]

### Step 6: Convert the volume of gas adsorbed to STP
Using the ideal gas law:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Where:
- \( P_1 = 400 \, \text{mmHg} \)
- \( V_1 = 125.67 \, \text{mL} \)
- \( T_1 = 300 \, \text{K} \) (27°C + 273)
- \( P_2 = 760 \, \text{mmHg} \) (STP)
- \( T_2 = 273 \, \text{K} \) (STP)

Rearranging to find \( V_2 \):
\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]
Substituting the values:
\[ V_2 = \frac{400 \, \text{mmHg} \times 125.67 \, \text{mL} \times 273 \, \text{K}}{760 \, \text{mmHg} \times 300 \, \text{K}} \]

Calculating:
\[ V_2 = \frac{400 \times 125.67 \times 273}{760 \times 300} \]
\[ V_2 = \frac{13840776}{228000} \approx 60.19 \, \text{mL} \]

### Final Answer
The volume of the gas adsorbed under the conditions of the experiment, reduced to STP, is approximately **60.19 mL**.

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