At `27^(@) C`, hydrogen is leaked through a tiny hole into a vessel for `20 min`. Another unknown gas at the same T and P as that of `H_(2)`, is leaked through the same hole for 20 min. After the effusion of the gases the mixture exerts a pressure of 6 atm. The hydrogen content of the mixture is 0.7 mole. If the volume of the container is 3 litre, what is molecular weight of unknown gas ?
`("Use" : R = 0.821 "L atm K"^(-1) "mole"^(-1))`

Options

A. 1088
B. 10.88
C. 108.8
D. 10880

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
- Temperature (T) = 27°C = 300 K (convert to Kelvin)
- Pressure of the mixture (P) = 6 atm
- Volume of the container (V) = 3 L
- Moles of hydrogen (n1) = 0.7 moles
- We need to find the moles of the unknown gas (n2) and its molecular weight.

### Step 2: Use the Ideal Gas Law to Find n2
The total pressure of the gas mixture can be expressed using the Ideal Gas Law:

\[ P = \frac{nRT}{V} \]

For the mixture of gases, we can write:

\[ P = P_1 + P_2 \]

Where:
- \( P_1 = \frac{n_1RT}{V} \)
- \( P_2 = \frac{n_2RT}{V} \)

Substituting these into the equation gives:

\[ \frac{n_1RT}{V} + \frac{n_2RT}{V} = 6 \]

Factoring out \( \frac{RT}{V} \):

\[ \frac{RT}{V}(n_1 + n_2) = 6 \]

### Step 3: Calculate n2
Now we can rearrange the equation to solve for \( n_2 \):

\[ n_1 + n_2 = \frac{6V}{RT} \]

Substituting the known values:

- \( n_1 = 0.7 \)
- \( V = 3 \, \text{L} \)
- \( R = 0.821 \, \text{L atm K}^{-1} \text{ mole}^{-1} \)
- \( T = 300 \, \text{K} \)

Calculating \( \frac{6V}{RT} \):

\[ \frac{6 \times 3}{0.821 \times 300} = \frac{18}{246.3} \approx 0.073 \, \text{moles} \]

Now, substituting back to find \( n_2 \):

\[ 0.7 + n_2 = 0.073 \implies n_2 = 0.073 - 0.7 = 0.0308 \, \text{moles} \]

### Step 4: Calculate the Rate of Effusion
The rate of effusion can be calculated as:

\[ \text{Rate} = \frac{\text{moles}}{\text{time}} \]

For hydrogen:

\[ \text{Rate of } H_2 = \frac{0.7}{20} \, \text{moles/min} \]

For the unknown gas:

\[ \text{Rate of unknown gas} = \frac{0.0308}{20} \, \text{moles/min} \]

### Step 5: Apply Graham's Law of Effusion
According to Graham's Law:

\[ \frac{\text{Rate of } H_2}{\text{Rate of unknown gas}} = \sqrt{\frac{M_{\text{unknown}}}{M_{H_2}}} \]

Where \( M_{H_2} = 2 \, \text{g/mol} \).

Substituting the rates:

\[ \frac{0.7/20}{0.0308/20} = \sqrt{\frac{M_{\text{unknown}}}{2}} \]

This simplifies to:

\[ \frac{0.7}{0.0308} = \sqrt{\frac{M_{\text{unknown}}}{2}} \]

### Step 6: Solve for Molar Mass of the Unknown Gas
Squaring both sides:

\[ \left(\frac{0.7}{0.0308}\right)^2 = \frac{M_{\text{unknown}}}{2} \]

Calculating the left side:

\[ \left(\frac{0.7}{0.0308}\right)^2 = \left(22.73\right)^2 \approx 516.5 \]

Now, substituting back to find \( M_{\text{unknown}} \):

\[ M_{\text{unknown}} = 2 \times 516.5 \approx 1033 \, \text{g/mol} \]

### Final Answer
The molecular weight of the unknown gas is approximately **1033 g/mol**.