For effusion of equal volume, a gas takes twice the time as taken by methane under similar conditions. The molar mass of the gas is

Options

A. 32 g/mol
B. 44 g/mol
C. 64 g/mol
D. 50 g/mol

Answer

To find the molar mass of the unknown gas based on the given conditions, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that the time taken for the effusion of the unknown gas is twice that of methane (CH₄). This means that if the time taken by methane is \( t \), the time taken by the unknown gas is \( 2t \).

2. **Relating Time and Rate of Effusion**:
The rate of effusion is inversely proportional to the time taken for effusion. Therefore, if the time taken by the unknown gas is twice that of methane, the rate of effusion of the unknown gas (let's call it \( R_1 \)) is half that of methane (let's call it \( R_2 \)):
\[ R_1 = \frac{R_2}{2} \]

3. **Applying Graham's Law**:
According to Graham's Law of Effusion, the rates of effusion of two gases are inversely proportional to the square root of their molar masses:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]
Where \( M_1 \) is the molar mass of the unknown gas and \( M_2 \) is the molar mass of methane.

4. **Substituting Known Values**:
We know that the molar mass of methane (CH₄) is \( 16 \, \text{g/mol} \). Substituting \( R_1 = \frac{R_2}{2} \) into Graham's Law gives:
\[ \frac{\frac{R_2}{2}}{R_2} = \sqrt{\frac{16}{M_1}} \]
This simplifies to:
\[ \frac{1}{2} = \sqrt{\frac{16}{M_1}} \]

5. **Squaring Both Sides**:
To eliminate the square root, we square both sides:
\[ \left(\frac{1}{2}\right)^2 = \frac{16}{M_1} \]
This simplifies to:
\[ \frac{1}{4} = \frac{16}{M_1} \]

6. **Cross-Multiplying to Solve for \( M_1 \)**:
Cross-multiplying gives:
\[ M_1 = 16 \times 4 \]
Therefore:
\[ M_1 = 64 \, \text{g/mol} \]

7. **Conclusion**:
The molar mass of the unknown gas is \( 64 \, \text{g/mol} \).

### Final Answer:
The molar mass of the gas is **64 g/mol**.