At what temperature will average speed of the molecules of the second member of the series `C_(n)H_(2n)` be the same to that of `Cl_(2)` at `627^(@)C`?

Options

A. 259.4 K
B. 400 K
C. 532.4 K
D. None of these

Explanation

To solve the problem, we need to find the temperature at which the average speed of the molecules of the second member of the series \( C_nH_{2n} \) (which is \( C_3H_6 \)) is equal to the average speed of \( Cl_2 \) at \( 627^\circ C \).

### Step-by-Step Solution:

1. **Identify the second member of the series \( C_nH_{2n} \)**:
- The first member (when \( n=2 \)) is \( C_2H_4 \).
- The second member (when \( n=3 \)) is \( C_3H_6 \).

2. **Calculate the molar mass of \( C_3H_6 \)**:
- Molar mass of Carbon (C) = 12 g/mol.
- Molar mass of Hydrogen (H) = 1 g/mol.
- Molar mass of \( C_3H_6 = (3 \times 12) + (6 \times 1) = 36 + 6 = 42 \) g/mol.

3. **Calculate the molar mass of \( Cl_2 \)**:
- Molar mass of Chlorine (Cl) = 35.5 g/mol.
- Molar mass of \( Cl_2 = 2 \times 35.5 = 71 \) g/mol.

4. **Convert the temperature of \( Cl_2 \) from Celsius to Kelvin**:
- Temperature in Celsius = \( 627^\circ C \).
- Temperature in Kelvin = \( 627 + 273 = 900 \) K.

5. **Use the formula for average speed of gas molecules**:
- The average speed \( v \) of gas molecules is given by:
\[ v = \sqrt{\frac{8RT}{\pi M}} \]
where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass in kg/mol.

6. **Set the average speeds equal**:
- For \( C_3H_6 \):
\[ v_{C_3H_6} = \sqrt{\frac{8RT}{\pi \cdot 0.042}} \]
- For \( Cl_2 \):
\[ v_{Cl_2} = \sqrt{\frac{8R \cdot 900}{\pi \cdot 0.071}} \]

7. **Equate the two speeds**:
\[ \sqrt{\frac{8RT}{\pi \cdot 0.042}} = \sqrt{\frac{8R \cdot 900}{\pi \cdot 0.071}} \]

8. **Square both sides to eliminate the square root**:
\[ \frac{8RT}{\pi \cdot 0.042} = \frac{8R \cdot 900}{\pi \cdot 0.071} \]

9. **Cancel out common terms**:
- \( 8R \) and \( \pi \) cancel out:
\[ \frac{T}{0.042} = \frac{900}{0.071} \]

10. **Cross-multiply to solve for \( T \)**:
\[ T \cdot 0.071 = 900 \cdot 0.042 \]
\[ T = \frac{900 \cdot 0.042}{0.071} \]

11. **Calculate \( T \)**:
\[ T = \frac{37.8}{0.071} \approx 532.4 \text{ K} \]

### Final Answer:
The temperature at which the average speed of the molecules of \( C_3H_6 \) will be the same as that of \( Cl_2 \) at \( 627^\circ C \) is approximately **532.4 K**.