50 mL of each gas A and of gas B takes 150 and 200 seconds respectively for effusing through a pin hole under the similar conditions. If molecular mass of gas A is 36, the molecular mass of gas B will be

Options

A. 96
B. 128
C. 32
D. 64

Explanation

To solve the problem, we will 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. The formula can be expressed as follows:

\[ \frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}} \]

Where:
- \( r_A \) and \( r_B \) are the rates of effusion of gases A and B, respectively.
- \( M_A \) and \( M_B \) are the molar masses of gases A and B, respectively.

### Step 1: Identify the given values
- Volume of gas A (\( V_A \)) = 50 mL
- Volume of gas B (\( V_B \)) = 50 mL
- Time taken for gas A (\( t_A \)) = 150 seconds
- Time taken for gas B (\( t_B \)) = 200 seconds
- Molar mass of gas A (\( M_A \)) = 36 g/mol

### Step 2: Calculate the rates of effusion
The rate of effusion can be defined as the volume of gas divided by the time taken for effusion. Since both gases have the same volume, we can express the rates as:

\[ r_A = \frac{V_A}{t_A} = \frac{50 \, \text{mL}}{150 \, \text{s}} = \frac{1}{3} \, \text{mL/s} \]

\[ r_B = \frac{V_B}{t_B} = \frac{50 \, \text{mL}}{200 \, \text{s}} = \frac{1}{4} \, \text{mL/s} \]

### Step 3: Set up the equation using Graham's law
Using Graham's law, we can set up the equation:

\[ \frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}} \]

Substituting the values we have:

\[ \frac{\frac{1}{3}}{\frac{1}{4}} = \sqrt{\frac{M_B}{36}} \]

### Step 4: Simplify the left side
Calculating the left side:

\[ \frac{1/3}{1/4} = \frac{4}{3} \]

So, we have:

\[ \frac{4}{3} = \sqrt{\frac{M_B}{36}} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:

\[ \left(\frac{4}{3}\right)^2 = \frac{M_B}{36} \]

Calculating the left side:

\[ \frac{16}{9} = \frac{M_B}{36} \]

### Step 6: Solve for \( M_B \)
Now, we can solve for \( M_B \):

\[ M_B = 36 \times \frac{16}{9} \]

Calculating \( M_B \):

\[ M_B = 36 \times \frac{16}{9} = 4 \times 16 = 64 \, \text{g/mol} \]

### Final Answer
The molecular mass of gas B is **64 g/mol**.

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