50 mL of gas A effuses through a pin hole in 146 seconds. The same volume of `CO_(2)` under identical condition effuses in 115 seconds. The molar mass of A is

Options

A. 44
B. 35.5
C. 71
D. None of these

Explanation

To find the molar mass of gas A using the information provided, we will apply 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.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Volume of gas A (V_A) = 50 mL
- Time taken for gas A to effuse (t_A) = 146 seconds
- Volume of CO₂ (V_CO₂) = 50 mL
- Time taken for CO₂ to effuse (t_CO₂) = 115 seconds

2. **Calculate the Rates of Effusion:**
- Rate of effusion for gas A (R_A) = V_A / t_A = 50 mL / 146 s
- Rate of effusion for CO₂ (R_CO₂) = V_CO₂ / t_CO₂ = 50 mL / 115 s

3. **Set Up the Equation Using Graham's Law:**
According to Graham's law:
\[ \frac{R_A}{R_{CO₂}} = \frac{\sqrt{M_{CO₂}}}{\sqrt{M_A}} \]
Rearranging gives:
\[ \frac{R_A}{R_{CO₂}} = \sqrt{\frac{M_{CO₂}}{M_A}} \]

4. **Substituting the Values:**
From the rates calculated:
\[ R_A = \frac{50}{146}, \quad R_{CO₂} = \frac{50}{115} \]
Thus,
\[ \frac{\frac{50}{146}}{\frac{50}{115}} = \sqrt{\frac{M_{CO₂}}{M_A}} \]
Simplifying gives:
\[ \frac{115}{146} = \sqrt{\frac{M_{CO₂}}{M_A}} \]

5. **Calculate Molar Mass of CO₂:**
The molar mass of CO₂ (M_CO₂) is:
\[ M_{CO₂} = 12 + 2 \times 16 = 44 \text{ g/mol} \]

6. **Square Both Sides to Eliminate the Square Root:**
\[ \left(\frac{115}{146}\right)^2 = \frac{44}{M_A} \]

7. **Cross Multiply to Solve for M_A:**
\[ M_A = 44 \times \left(\frac{146}{115}\right)^2 \]

8. **Calculate the Value:**
First, calculate \(\left(\frac{146}{115}\right)^2\):
\[ \frac{146}{115} \approx 1.2696 \quad \Rightarrow \quad (1.2696)^2 \approx 1.613 \]
Now, multiply by 44:
\[ M_A \approx 44 \times 1.613 \approx 71.0 \text{ g/mol} \]

### Final Answer:
The molar mass of gas A is approximately **71 g/mol**.