Molar entropy change is 16 J `mol^(-1)K^(-1)` , the boiling points of the liquid is if molar heat of vaporization is 6 kJ/mol.

Options

A. `375^(@)` C
B. 375 K
C. 273 K
D. `102^(@)` C

Explanation

To find the boiling point of the liquid given the molar entropy change and the molar heat of vaporization, we can use the relationship between entropy change (ΔS), heat of vaporization (ΔH), and temperature (T) at the boiling point. The formula we will use is:

\[ \Delta S = \frac{\Delta H}{T} \]

### Step-by-Step Solution:

1. **Identify the given values:**
- Molar entropy change (ΔS) = 16 J/mol·K
- Molar heat of vaporization (ΔH) = 6 kJ/mol = 6000 J/mol (since 1 kJ = 1000 J)

2. **Set up the equation using the formula:**
\[ \Delta S = \frac{\Delta H}{T} \]
Rearranging the formula to solve for T gives:
\[ T = \frac{\Delta H}{\Delta S} \]

3. **Substitute the values into the equation:**
\[ T = \frac{6000 \, \text{J/mol}}{16 \, \text{J/mol·K}} \]

4. **Calculate the temperature:**
\[ T = \frac{6000}{16} = 375 \, \text{K} \]

5. **Conclusion:**
The boiling point of the liquid is 375 K.