A solution of `122gm` of benzoic acid is `1000gm` of benzene shows a `b.p.` elevation of `1.4^(@)`. Assuming that solute is dimerized to the extent of `80` percent `(80^(@)C)` calculate normal `b.p.` of benzene given molar enthalpy of vap. Of benzine`=7.8 Kcal//mol`.

Explanation

To solve the problem of calculating the normal boiling point of benzene given the boiling point elevation due to benzoic acid, we can follow these steps:

### Step 1: Identify Given Data
- Mass of benzoic acid (solute) = 122 g
- Mass of benzene (solvent) = 1000 g
- Boiling point elevation (ΔTb) = 1.4 °C
- Percentage of dimerization (α) = 80% = 0.8
- Molar enthalpy of vaporization of benzene = 7.8 kcal/mol
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