A $0.50 m$ aqueous $NaCl$ solution has an experimentally determined van't Hoff factor $i = 1.85$ . With $K_{b} = 0.52 K kg mo l^{- 1}$ and the normal boiling point of water $100.00^{\circ} C$ , find the boiling point of the solution in $^{\circ} C$ to two decimal places.

Numerical answer type, answer in degC, tolerance 0.01.

Answer: 100.48 degC

Δ T_{b} = i K_{b} m = 1.85 \times 0.52 \times 0.50 = 0.481 K

, so the boiling point is

100.00 + 0.481 = 100.48^{\circ} C