A certain gas takes three times as long to effuse out as helium. Its molar mass will be

Options

A. `27 u`
B. `36 u`
C. `64 u`
D. `9 u`

Explanation

To find the molar mass of the certain gas that takes three times as long to effuse as helium, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We know that the time taken for the gas to effuse (T_gas) is three times the time taken for helium to effuse (T_He). Therefore, we can write:
\[ T_{\text{gas}} = 3 \times T_{\text{He}} \]

2. **Graham's Law of Effusion**: According to Graham's law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This can be expressed as:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]
where \(R_1\) and \(R_2\) are the rates of effusion of gas 1 and gas 2, and \(M_1\) and \(M_2\) are their respective molar masses.

3. **Relate Time and Rate**: The rate of effusion is inversely related to the time of effusion. Therefore, we can express the rates in terms of time:
\[ R_{\text{gas}} = \frac{1}{T_{\text{gas}}} \quad \text{and} \quad R_{\text{He}} = \frac{1}{T_{\text{He}}} \]
Thus, we can rewrite Graham's law as:
\[ \frac{T_{\text{He}}}{T_{\text{gas}}} = \sqrt{\frac{M_{\text{gas}}}{M_{\text{He}}}} \]

4. **Substituting Known Values**: We know that \(T_{\text{gas}} = 3 \times T_{\text{He}}\). Substituting this into the equation gives:
\[ \frac{T_{\text{He}}}{3 \times T_{\text{He}}} = \sqrt{\frac{M_{\text{gas}}}{M_{\text{He}}}} \]
Simplifying this, we get:
\[ \frac{1}{3} = \sqrt{\frac{M_{\text{gas}}}{M_{\text{He}}}} \]

5. **Squaring Both Sides**: To eliminate the square root, we square both sides:
\[ \left(\frac{1}{3}\right)^2 = \frac{M_{\text{gas}}}{M_{\text{He}}} \]
This simplifies to:
\[ \frac{1}{9} = \frac{M_{\text{gas}}}{M_{\text{He}}} \]

6. **Finding Molar Mass of Helium**: The molar mass of helium (M_He) is known to be 4 g/mol. Therefore, we can substitute this value into the equation:
\[ \frac{1}{9} = \frac{M_{\text{gas}}}{4} \]

7. **Solving for Molar Mass of the Gas**: Rearranging the equation to solve for \(M_{\text{gas}}\):
\[ M_{\text{gas}} = 4 \times \frac{1}{9} = \frac{4}{9} \text{ g/mol} \]
But we need to find the molar mass in terms of the square of the time ratio:
\[ M_{\text{gas}} = 4 \times 3^2 = 4 \times 9 = 36 \text{ g/mol} \]

8. **Final Answer**: Thus, the molar mass of the gas is:
\[ M_{\text{gas}} = 36 \text{ g/mol} \]