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, we will follow these steps:

### Step 1: Calculate the density of the gas mixture
The density of the gas mixture is given as:
\[ \text{Density} = \frac{64}{246.3} \text{ gm/litre} \]

### Step 2: Use the ideal gas equation to find the molecular mass of the mixture
We can use the formula relating pressure, density, and molecular mass:
\[ P = \frac{dRT}{M} \]
Rearranging gives us:
\[ M = \frac{dRT}{P} \]
Where:
- \( d = \frac{64}{246.3} \) gm/litre
- \( R = 0.0821 \) L·atm/(K·mol)
- \( T = 300 \) K
- \( P = 1 \) atm

Substituting the values:
\[ M = \frac{\left(\frac{64}{246.3}\right) \times (0.0821) \times (300)}{1} \]

### Step 3: Calculate the molecular mass
Calculating the above expression:
\[ M = \frac{64 \times 0.0821 \times 300}{246.3} \]
Calculating this gives:
\[ M \approx 6.4 \text{ g/mol} \]

### Step 4: Relate the molecular mass to the composition of the gas mixture
The molecular mass of the mixture can also be expressed in terms of the number of moles of helium (\( n_{He} \)) and methane (\( n_{CH_4} \)):
\[ M = \frac{n_{He} \cdot M_{He} + n_{CH_4} \cdot M_{CH_4}}{n_{He} + n_{CH_4}} \]
Where:
- \( M_{He} = 4 \) g/mol (molecular mass of helium)
- \( M_{CH_4} = 16 \) g/mol (molecular mass of methane)

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

### Step 5: Set up the equation
Multiplying both sides by \( (n_{He} + n_{CH_4}) \):
\[ 6.4(n_{He} + n_{CH_4}) = 4n_{He} + 16n_{CH_4} \]
Expanding this gives:
\[ 6.4n_{He} + 6.4n_{CH_4} = 4n_{He} + 16n_{CH_4} \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ 2.4n_{He} = 9.6n_{CH_4} \]
Dividing both sides by \( n_{CH_4} \) gives:
\[ \frac{n_{He}}{n_{CH_4}} = \frac{9.6}{2.4} = 4 \]

### Step 7: Conclusion
Thus, the composition of the gas mixture effusing out initially is:
\[ n_{He} : n_{CH_4} = 4 : 1 \]