The correct relationship between enthalpy (in cal `mol^(-1)`) of vaporisation of liquid and its boiling point is

Options

A. `Delta H_("vap") ~~ 21 T_(b)`
B. `Delta H_("vap") ~~ 1.55 T_(b)`
C. `Delta H_("vap") ~~ 2//3 T_(b)`
D. `Delta H_("vap") ~~ 2.303 T_(b)`

Explanation

To solve the question regarding the relationship between the enthalpy of vaporization of a liquid and its boiling point, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Enthalpy of Vaporization**:
- The enthalpy of vaporization (ΔH_vaporization) is the amount of energy required to convert a unit quantity of a liquid into vapor at its boiling point.

2. **Introduce Trouton's Law**:
- Trouton's Law states that the enthalpy of vaporization of a liquid is approximately proportional to its boiling point temperature (T_B). The relationship can be expressed mathematically.

3. **Mathematical Expression**:
- According to Trouton's Law:
\[ \frac{\Delta H_{vaporization}}{T_B} = 21 \, \text{cal} \, \text{K}^{-1} \, \text{mol}^{-1} \]
- Here, ΔH_vaporization is in calories per mole and T_B is the boiling point in Kelvin.

4. **Rearranging the Equation**:
- To find the enthalpy of vaporization, we can rearrange the equation:
\[ \Delta H_{vaporization} = 21 \times T_B \]

5. **Conclusion**:
- Therefore, the relationship between the enthalpy of vaporization and the boiling point of a liquid can be summarized as:
\[ \Delta H_{vaporization} = 21 \, \text{cal} \, \text{K}^{-1} \, \text{mol}^{-1} \times T_B \]
- This indicates that as the boiling point increases, the enthalpy of vaporization also increases.

### Final Answer:
The correct relationship is given by Trouton's Law:
\[ \Delta H_{vaporization} = 21 \times T_B \]