Calculate the entropy change of n-hexane when 1 mole of it evaporates at 341.7 K`(Delta_(vap) H = 29.0 kJ "mol"^(-1))`.

Explanation

To calculate the entropy change (ΔS) of n-hexane when 1 mole of it evaporates at 341.7 K, we can use the formula:

\[ \Delta S = \frac{\Delta H_{vap}}{T} \]

where:
- ΔS is the change in entropy,
- ΔH_{vap} is the heat of vaporization,
- T is the temperature in Kelvin.

### Step-by-Step Solution:

1. **Identify the values**:
- ΔH_{vap} = 29.0 kJ/mol
- T = 341.7 K

2. **Convert ΔH_{vap} from kJ to J**:
\[ \Delta H_{vap} = 29.0 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 29000 \, \text{J/mol} \]

3. **Substitute the values into the entropy change formula**:
\[ \Delta S = \frac{29000 \, \text{J/mol}}{341.7 \, \text{K}} \]

4. **Calculate ΔS**:
\[ \Delta S = \frac{29000}{341.7} \approx 84.87 \, \text{J/K/mol} \]

5. **Final Result**:
The entropy change (ΔS) when 1 mole of n-hexane evaporates at 341.7 K is approximately:
\[ \Delta S \approx 84.87 \, \text{J/K/mol} \]