The entropy change for vaporisation of a liquid is `109.3JK^(-1)mol^(-1)`. The molar heat of vaporisation of that liquid is 40.77 kJ `mol^(-1)`. Calculate the boiling point of that liquid.

Explanation

To solve the problem, we need to calculate the boiling point of the liquid using the given values of entropy change and molar heat of vaporization. We'll use the relationship between entropy change (ΔS), heat of vaporization (ΔHvap), and temperature (T) at equilibrium.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Entropy change for vaporization (ΔS) = 109.3 J K⁻¹ mol⁻¹
- Molar heat of vaporization (ΔHvap) = 40.77 kJ mol⁻¹

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