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, we can use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understand the Problem**:
We know that the unknown gas takes twice the time to effuse compared to methane (CH₄). If the time taken by methane is T, then the time taken by the unknown gas is 2T.

2. **Apply Graham's Law**:
According to Graham's law, the rates of effusion of two gases are inversely proportional to the square roots of their molar masses. We can express this relationship as:
\[ \frac{\text{Rate of CH}_4}{\text{Rate of unknown gas}} = \frac{\sqrt{M_{\text{unknown}}}}{\sqrt{M_{\text{CH}_4}}} \]

3. **Relate Rate and Time**:
The rate of effusion is inversely proportional to the time taken. Therefore, we can write:
\[ \frac{\text{Rate of CH}_4}{\text{Rate of unknown gas}} = \frac{2T}{T} = 2 \]

4. **Set Up the Equation**:
Now we can set up the equation using the relationship from Graham's law:
\[ 2 = \frac{\sqrt{M_{\text{unknown}}}}{\sqrt{M_{\text{CH}_4}}} \]

5. **Substitute the Molar Mass of Methane**:
The molar mass of methane (CH₄) is 16 g/mol. Substituting this into the equation gives:
\[ 2 = \frac{\sqrt{M_{\text{unknown}}}}{\sqrt{16}} \]

6. **Simplify the Equation**:
Simplifying the right side:
\[ 2 = \frac{\sqrt{M_{\text{unknown}}}}{4} \]

7. **Solve for Molar Mass of the Unknown Gas**:
Multiply both sides by 4:
\[ 8 = \sqrt{M_{\text{unknown}}} \]
Now, square both sides to find \(M_{\text{unknown}}\):
\[ M_{\text{unknown}} = 8^2 = 64 \text{ g/mol} \]

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

---