For 10 minutes each, at `0^(@)C`, from two identical holes nitrogen and an unknown gas are leaked into a common vessel of 4 litre capacity. The resulting pressure is 2.8 atm and the mixture contains 0.4 mole of nitrogen. What is the molar mass of unknown gas ? (Take `R = 0.821 "L-atm mol"^(-1)K^(-1)`]

Options

A. `448 g mol^(-1)`
B. `224 g mol^(-1)`
C. `226 g mol^(-1)`
D. None of these

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the Given Data
- Volume of the vessel (V) = 4 L
- Total pressure (P) = 2.8 atm
- Moles of nitrogen (N₂) = 0.4 moles
- Temperature (T) = 0°C = 273 K (convert to Kelvin)
- Ideal gas constant (R) = 0.821 L·atm/(mol·K)

### Step 2: Calculate Total Moles of Gas in the Vessel
Using the ideal gas equation:
\[ PV = nRT \]
Where:
- \( n \) = total number of moles of gas
- Rearranging gives:
\[ n = \frac{PV}{RT} \]

Substituting the known values:
\[ n = \frac{(2.8 \, \text{atm}) \times (4 \, \text{L})}{(0.821 \, \text{L·atm/(mol·K)}) \times (273 \, \text{K})} \]

Calculating:
\[ n = \frac{11.2}{224.793} \approx 0.0498 \, \text{moles} \]

### Step 3: Calculate Moles of Unknown Gas
Since we know the total moles and the moles of nitrogen:
\[ n_{\text{unknown}} = n_{\text{total}} - n_{\text{N}_2} \]
\[ n_{\text{unknown}} = 0.5 - 0.4 = 0.1 \, \text{moles} \]

### Step 4: Use Graham's Law of Effusion
According to Graham's law:
\[ \frac{R_{N_2}}{R_{unknown}} = \sqrt{\frac{M_{unknown}}{M_{N_2}}} \]
Where:
- \( R_{N_2} = \frac{0.4 \, \text{moles}}{10 \, \text{minutes}} = 0.04 \, \text{moles/min} \)
- \( R_{unknown} = \frac{0.1 \, \text{moles}}{10 \, \text{minutes}} = 0.01 \, \text{moles/min} \)
- Molar mass of nitrogen (\( M_{N_2} \)) = 28 g/mol

### Step 5: Substitute Values into Graham's Law
Substituting the values into Graham's law:
\[ \frac{0.04}{0.01} = \sqrt{\frac{M_{unknown}}{28}} \]
\[ 4 = \sqrt{\frac{M_{unknown}}{28}} \]

### Step 6: Square Both Sides
Squaring both sides gives:
\[ 16 = \frac{M_{unknown}}{28} \]

### Step 7: Solve for Molar Mass of Unknown Gas
Rearranging gives:
\[ M_{unknown} = 16 \times 28 = 448 \, \text{g/mol} \]

### Final Answer
The molar mass of the unknown gas is **448 g/mol**.

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