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 enthalpy change (ΔH), entropy change (ΔS), and temperature (T). The relevant formula is:

\[ \Delta H = T \Delta S \]

### 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

2. **Convert ΔH to Joules:**
Since ΔH is given in kJ/mol, we need to convert it to J/mol for consistency:
\[ \Delta H = 6 \text{ kJ/mol} \times 1000 \text{ J/kJ} = 6000 \text{ J/mol} \]

3. **Rearrange the Formula:**
We need to solve for T (boiling temperature):
\[ T = \frac{\Delta H}{\Delta S} \]

4. **Substitute the Values:**
Now, substitute the values of ΔH and ΔS into the equation:
\[ T = \frac{6000 \text{ J/mol}}{16 \text{ J/mol·K}} \]

5. **Calculate T:**
Performing the division:
\[ T = \frac{6000}{16} = 375 \text{ K} \]

6. **Final Answer:**
The boiling point of the liquid is 375 K.