1 mole of `H_(2)O` is vaporised at `100^(@)C` and 1 atm pressure. If the latent heat of vaporisation of water is x J/g, then `Delta S` in terms of J/mole K is given by

Options

A. `(x)/(373)`
B. `(18x)/(300)`
C. `(18x)/(373)`
D. `(1)/(2)xx(18x)/(373)`

Answer

To solve the problem, we need to calculate the change in entropy (ΔS) when 1 mole of water (H₂O) is vaporized at 100°C and 1 atm pressure. The latent heat of vaporization is given as x J/g.

### Step-by-Step Solution:

1. **Identify the Latent Heat of Vaporization**:
- The latent heat of vaporization (L) for water is given as x J/g.
- For 1 mole of water, which has a molar mass of 18 g, the total heat required (Q) for vaporization is:
\[ Q = 18 \, \text{g} \times x \, \text{J/g} = 18x \, \text{J} \]

2. **Convert Temperature to Kelvin**:
- The temperature at which the vaporization occurs is given as 100°C. To convert this to Kelvin:
\[ T = 100 + 273 = 373 \, \text{K} \]

3. **Calculate the Change in Entropy (ΔS)**:
- The change in entropy (ΔS) during a phase transition can be calculated using the formula:
\[ \Delta S = \frac{Q}{T} \]
- Substituting the values we found:
\[ \Delta S = \frac{18x \, \text{J}}{373 \, \text{K}} = \frac{18x}{373} \, \text{J/mol K} \]

4. **Final Expression for ΔS**:
- Therefore, the change in entropy (ΔS) for the vaporization of 1 mole of water is:
\[ \Delta S = \frac{18x}{373} \, \text{J/mol K} \]

### Conclusion:
The final answer for ΔS in terms of J/mol K is:
\[ \Delta S = \frac{18x}{373} \, \text{J/mol K} \]