An aqueous solution of a non-volatile solute freezes at 272.4 K, while pure water freezes at 273.0 K. Determine the following:
(Given `K_(f)= 1.86 "K kg mol"^(-1), K_(b) = 0.512 "K kg mol"^(-1)` and vapour pressure of water at 298 K = 23.756 mm of Hg)
(1) The molality of solution
(2) Boiling point of solution
(3) The lowering of vapour pressure of water at 298 K

Explanation

To solve the problem, we will break it down into three parts as specified in the question.

### Step 1: Calculate the Molality of the Solution

1. **Determine the depression in freezing point (ΔTf)**:
\[ \Delta T_f = T_f(\text{pure water}) - T_f(\text{solution}) = 273.0 \, \text{K} - 272.4 \, \text{K} = 0.6 \, \text{K} \]
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