Molar heat of vaporization of a liquid is 6 `KJ mol^(-1)`. If its entropy change is `16 JK^(-1)mol^(-1)`, then boiling point of the liquid is

Options

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

Answer

To find the boiling point of the liquid given the molar heat of vaporization and the entropy change, we can use the relationship between these quantities. The formula we will use is:

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

Where:
- \(\Delta S\) is the entropy change,
- \(\Delta H\) is the heat of vaporization,
- \(T\) is the temperature in Kelvin.

### Step 1: Convert the heat of vaporization from kJ/mol to J/mol
The molar heat of vaporization is given as 6 kJ/mol. To convert this to joules, we multiply by 1000:

\[ \Delta H = 6 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 6000 \, \text{J/mol} \]

### Step 2: Use the formula to find the boiling point
Now we can rearrange the formula to solve for \(T\):

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

Substituting the values we have:

\[ T = \frac{6000 \, \text{J/mol}}{16 \, \text{J/K/mol}} = 375 \, \text{K} \]

### Conclusion
The boiling point of the liquid is 375 K.

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