Calculate the entropy change involved in the vaporisation of water at 373 K to vapours at the same temperature(Latent heat of vaporisation = 2.275 kJ/g).

Explanation

To calculate the entropy change involved in the vaporization of water at 373 K, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Latent heat of vaporization (L) = 2.275 kJ/g
- Temperature (T) = 373 K
- Molar mass of water (H₂O) = 18 g/mol

2. **Convert Latent Heat to Molar Enthalpy:**
- First, we need to convert the latent heat of vaporization from kJ/g to kJ/mol.
- Molar enthalpy of vaporization (ΔH) can be calculated as:
\[ \Delta H = L \times \text{Molar mass of water} = 2.275 \, \text{kJ/g} \times 18 \, \text{g/mol} \]
- Calculate:
\[ \Delta H = 2.275 \times 18 = 40.95 \, \text{kJ/mol} \]

3. **Convert Molar Enthalpy to Joules:**
- Since we need the entropy change in Joules, convert ΔH from kJ/mol to J/mol:
\[ \Delta H = 40.95 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 40950 \, \text{J/mol} \]

4. **Calculate the Entropy Change (ΔS):**
- The entropy change during vaporization can be calculated using the formula:
\[ \Delta S = \frac{\Delta H}{T} \]
- Substitute the values:
\[ \Delta S = \frac{40950 \, \text{J/mol}}{373 \, \text{K}} \]
- Calculate:
\[ \Delta S \approx 109.78 \, \text{J/K·mol} \]

5. **Final Result:**
- The entropy change involved in the vaporization of water at 373 K is approximately:
\[ \Delta S \approx 109.78 \, \text{J/K·mol} \]