Sea water contains $$29.25$$% NaCl and $$19$$% MgCl$$_2$$ by weight of solution. The normal boiling point of the sea water is _____ °C (Nearest integer) Assume 100% ionization for both NaCl and MgCl$$_2$$. Given:
K$$_b$$H$$_2$$O $$= 0.52$$ K kg mol$$^{-1}$$. Molar mass of NaCl and MgCl$$_2$$ is $$58.5$$ and $$95$$ g mol$$^{-1}$$ respectively.

Given: 29.25% NaCl and 19% MgCl₂ by weight. K_b = 0.52 K kg/mol.

Consider 100 g of sea water: 29.25 g NaCl, 19 g MgCl₂, solvent = 100 - 29.25 - 19 = 51.75 g = 0.05175 kg.

Moles: NaCl = 29.25/58.5 = 0.5 mol, MgCl₂ = 19/95 = 0.2 mol.

Van't Hoff factors: NaCl → Na⁺ + Cl⁻ (i = 2), MgCl₂ → Mg²⁺ + 2Cl⁻ (i = 3).

Total molality of particles: $$\frac{0.5 \times 2 + 0.2 \times 3}{0.05175} = \frac{1.6}{0.05175} = 30.918$$ mol/kg

$$\Delta T_b = K_b \times m = 0.52 \times 30.918 = 16.08$$ °C

Boiling point = 100 + 16.08 ≈ 116 °C

The answer is $$\boxed{116}$$ °C.