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

Options

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

Explanation

To solve the problem of finding the molar mass of a gas that takes three times as long to effuse as helium, we can use Graham's law of effusion. Here's a step-by-step solution:

### Step 1: Understand Graham's Law of Effusion
Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it 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, respectively.
- \( M_1 \) and \( M_2 \) are the molar masses of gas 1 and gas 2, respectively.

### Step 2: Define the Variables
Let:
- Gas 1 be helium (He) with a molar mass \( M_1 = 4 \, \text{g/mol} \).
- Gas 2 be the unknown gas with molar mass \( M_2 \).
- The time taken for helium to effuse be \( T_1 \).
- The time taken for the unknown gas to effuse be \( T_2 = 3T_1 \) (since it takes three times longer).

### Step 3: Relate Rates of Effusion to Time
The rate of effusion is inversely related to the time taken. Therefore, we can express the rates as:

\[ r_1 = \frac{V}{T_1} \quad \text{and} \quad r_2 = \frac{V}{T_2} \]

Thus, we can write:

\[ \frac{r_1}{r_2} = \frac{T_2}{T_1} = \frac{3T_1}{T_1} = 3 \]

### Step 4: Substitute into Graham's Law
Now substituting the values into Graham's law:

\[ 3 = \sqrt{\frac{M_2}{4}} \]

### Step 5: Square Both Sides
To eliminate the square root, square both sides:

\[ 3^2 = \frac{M_2}{4} \]

This simplifies to:

\[ 9 = \frac{M_2}{4} \]

### Step 6: Solve for Molar Mass \( M_2 \)
Now, multiply both sides by 4 to find \( M_2 \):

\[ M_2 = 9 \times 4 = 36 \, \text{g/mol} \]

### Final Answer
The molar mass of the gas is \( 36 \, \text{g/mol} \).

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