An aqueous solution freezes at 271.5 K. Determine its boiling point and vapour pressure at 298 K. The cryoscopic constant of water is `1.86^@`, its ebullioscopic constant is 0.5160 and the water vapour pressure at 298 K is 3168 Pa.

Explanation

To solve the problem, we need to determine the boiling point and vapor pressure of an aqueous solution that freezes at 271.5 K. We will use the cryoscopic and ebullioscopic constants of water, as well as the vapor pressure of pure water at 298 K.

### Step 1: Calculate the Depression in Freezing Point (ΔTf)

The freezing point of pure water is 273 K. The freezing point of the solution is given as 271.5 K.

\[
\Delta T_f = T_f^{\text{pure}} - T_f^{\text{solution}} = 273 \, \text{K} - 271.5 \, \text{K} = 1.5 \, \text{K}
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