Under identical conditions of pressure and temperature, 4 L of gaseous moxture (`H_(2)` and `CH_(4)`) effuses through a hole in 5 min whereas 4 L of a gas X of molecular mass 36 takes to 10 min to effuse through the same hole. The mole ratio of `H_(2):CH_(4)` in the mixture is:

Options

A. `1:2`
B. `2:1`
C. `2:3`
D. `1:1`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between effusion rates and molecular mass
The rate of effusion of a gas is inversely proportional to the square root of its molecular mass. This can be expressed mathematically as:
\[ \text{Rate} \propto \frac{1}{\sqrt{M}} \]
where \( M \) is the molecular mass of the gas.

### Step 2: Calculate the effusion rates
Given:
- The volume of the gaseous mixture (H₂ and CH₄) is 4 L, and it effuses in 5 minutes.
- The volume of gas X (with a molecular mass of 36 g/mol) is also 4 L, and it effuses in 10 minutes.

The rate of effusion for the mixture can be calculated as:
\[ \text{Rate of mixture} = \frac{4 \, \text{L}}{5 \, \text{min}} = 0.8 \, \text{L/min} \]

The rate of effusion for gas X can be calculated as:
\[ \text{Rate of gas X} = \frac{4 \, \text{L}}{10 \, \text{min}} = 0.4 \, \text{L/min} \]

### Step 3: Set up the ratio of effusion rates
Using the relationship between the rates of effusion and molecular masses, we can write:
\[ \frac{\text{Rate of mixture}}{\text{Rate of gas X}} = \sqrt{\frac{M_X}{M_{\text{mixture}}}} \]

Substituting the values we calculated:
\[ \frac{0.8}{0.4} = \sqrt{\frac{36}{M_{\text{mixture}}}} \]
This simplifies to:
\[ 2 = \sqrt{\frac{36}{M_{\text{mixture}}}} \]

### Step 4: Solve for the molecular mass of the mixture
Squaring both sides gives:
\[ 4 = \frac{36}{M_{\text{mixture}}} \]
Rearranging this equation, we find:
\[ M_{\text{mixture}} = \frac{36}{4} = 9 \, \text{g/mol} \]

### Step 5: Express the molecular mass of the mixture in terms of mole fractions
Let \( x_{H_2} \) be the mole fraction of H₂ and \( x_{CH_4} \) be the mole fraction of CH₄. The molecular masses are:
- H₂: 2 g/mol
- CH₄: 16 g/mol

The molecular mass of the mixture can be expressed as:
\[ M_{\text{mixture}} = (2 \, \text{g/mol} \cdot x_{H_2}) + (16 \, \text{g/mol} \cdot x_{CH_4}) \]
Since \( x_{CH_4} = 1 - x_{H_2} \), we can write:
\[ M_{\text{mixture}} = 2x_{H_2} + 16(1 - x_{H_2}) = 2x_{H_2} + 16 - 16x_{H_2} = 16 - 14x_{H_2} \]

### Step 6: Set the equation for molecular mass equal to 9
Now, we set the equation equal to the molecular mass we found:
\[ 9 = 16 - 14x_{H_2} \]
Rearranging gives:
\[ 14x_{H_2} = 16 - 9 = 7 \]
Thus, we find:
\[ x_{H_2} = \frac{7}{14} = \frac{1}{2} \]

### Step 7: Calculate the mole fraction of CH₄
Since \( x_{CH_4} = 1 - x_{H_2} \):
\[ x_{CH_4} = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 8: Determine the mole ratio of H₂ to CH₄
The mole ratio of H₂ to CH₄ is:
\[ \text{Mole ratio} = \frac{x_{H_2}}{x_{CH_4}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1:1 \]

### Final Answer
The mole ratio of H₂ to CH₄ in the mixture is **1:1**.

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