A gaseous mixture `(He and CH_(4))` which has density `(64)/(246.3)` gm/litre at 1 atm & 300 K is kept in a container. Now a pinhole is made on the wall of the container through which He(g) and `CH_(4)` effuses. What will be the composition of the gas mixture `[n_(He) : n_(CH_(4))]` effusing out initially?

Options

A. `4 : 1`
B. `8 : 1`
C. `2 : 1`
D. `16 : 1`

Explanation

To solve the problem of determining the composition of the gas mixture (He and CH₄) that effuses out initially through a pinhole, we can follow these steps:

### Step 1: Calculate the Molar Mass of the Gas Mixture
Using the ideal gas equation, we can express the molar mass of the gas mixture in terms of its density, temperature, and pressure.

The ideal gas equation is given by:
\[ PV = nRT \]

Where:
- \( P \) = pressure (1 atm)
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (300 K)

We can express the number of moles \( n \) as:
\[ n = \frac{mass}{molar\ mass} \]

Substituting this into the ideal gas equation gives:
\[ P = \frac{density \times molar\ mass}{RT} \]

Rearranging for molar mass:
\[ molar\ mass = \frac{density \times RT}{P} \]

Given:
- Density = \( \frac{64}{246.3} \) g/L
- \( R = 0.0821 \) L·atm/(K·mol)
- \( T = 300 \) K
- \( P = 1 \) atm

Now substituting the values:
\[ molar\ mass = \frac{\left(\frac{64}{246.3}\right) \times (0.0821) \times (300)}{1} \]

Calculating this gives:
\[ molar\ mass \approx 6.4 \text{ g/mol} \]

### Step 2: Determine the Molar Mass of Each Component
The molar masses of the components are:
- Helium (He) = 4 g/mol
- Methane (CH₄) = 16 g/mol

### Step 3: Set Up the Mass Balance Equation
Let \( n_{He} \) and \( n_{CH_4} \) be the number of moles of He and CH₄ in the mixture, respectively. The average molar mass of the mixture can be expressed as:
\[ \frac{n_{He} \cdot M_{He} + n_{CH_4} \cdot M_{CH_4}}{n_{He} + n_{CH_4}} = 6.4 \]

Substituting the molar masses:
\[ \frac{n_{He} \cdot 4 + n_{CH_4} \cdot 16}{n_{He} + n_{CH_4}} = 6.4 \]

### Step 4: Solve for the Ratio of Moles
Cross-multiplying gives:
\[ n_{He} \cdot 4 + n_{CH_4} \cdot 16 = 6.4(n_{He} + n_{CH_4}) \]

Expanding and rearranging:
\[ 4n_{He} + 16n_{CH_4} = 6.4n_{He} + 6.4n_{CH_4} \]
\[ 4n_{He} - 6.4n_{He} + 16n_{CH_4} - 6.4n_{CH_4} = 0 \]
\[ -2.4n_{He} + 9.6n_{CH_4} = 0 \]

This simplifies to:
\[ \frac{n_{He}}{n_{CH_4}} = \frac{9.6}{2.4} = 4 \]

### Step 5: Final Ratio of Moles
Thus, the ratio of moles of helium to methane in the effusing gas mixture is:
\[ n_{He} : n_{CH_4} = 4 : 1 \]

### Conclusion
The composition of the gas mixture effusing out initially is \( 4 : 1 \).

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