At what temperature, the rate of effusion of `N_(2)` would be 1.625 times that of `SO_(2)` at `50^(@)C` ?

Options

A. 110K
B. 173K
C. 373 K
D. 273K

Answer

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 and directly proportional to the square root of the temperature.

### Step-by-step Solution:

1. **Identify the Given Information:**
- We need to find the temperature \( T_1 \) at which the rate of effusion of \( N_2 \) is 1.625 times that of \( SO_2 \).
- The temperature \( T_2 \) is given as \( 50^\circ C \) which we will convert to Kelvin.

2. **Convert Celsius to Kelvin:**
\[ T_2 = 50 + 273 = 323 \, K \]

3. **Write the Formula Based on Graham's Law:**
According to Graham's law:
\[ \frac{R_1}{R_2} = \sqrt{\frac{T_1}{T_2}} \cdot \frac{M_2}{M_1} \]
Where:
- \( R_1 \) is the rate of effusion of \( N_2 \)
- \( R_2 \) is the rate of effusion of \( SO_2 \)
- \( M_1 \) is the molar mass of \( N_2 \) (28 g/mol)
- \( M_2 \) is the molar mass of \( SO_2 \) (64 g/mol)

4. **Substituting Known Values:**
We know \( R_1 = 1.625 R_2 \):
\[ 1.625 = \sqrt{\frac{T_1}{323}} \cdot \frac{64}{28} \]

5. **Rearranging the Equation:**
\[ \sqrt{\frac{T_1}{323}} = \frac{1.625 \cdot 28}{64} \]

6. **Calculating the Right Side:**
\[ \frac{1.625 \cdot 28}{64} = \frac{45.5}{64} \approx 0.7109 \]

7. **Squaring Both Sides:**
\[ \frac{T_1}{323} = (0.7109)^2 \]
\[ \frac{T_1}{323} \approx 0.5053 \]

8. **Solving for \( T_1 \):**
\[ T_1 \approx 0.5053 \cdot 323 \approx 162.3 \, K \]

9. **Final Calculation:**
\[ T_1 = 1.625^2 \cdot \frac{28}{64} \cdot 323 \]
\[ T_1 \approx 373.15 \, K \]

### Conclusion:
The temperature at which the rate of effusion of \( N_2 \) would be 1.625 times that of \( SO_2 \) at \( 50^\circ C \) is approximately **373 K**.