Seawater contains 29.25% NaCl and 47.5% MgCl₂by weight of solution. The normal boiling point of the seawater is ________ ^oC (Nearest integer)
Assume 50% ionization for both NaCl and MgCl₂
Given: K_b (H₂O) 0.532K kg mol^-1
The molar mass of NaCl and MgCl₂ is 58.5 and 95 g mol^-1 respectively.

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413

40.04

140

100.404

answer is C.

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Detailed Solution

Number of moles of NaCl = $\frac{29.25}{58.5} \times [1 + 0.5 (2 - 1)] = 0.75$

Number of moles of MgCl₂ = $\frac{47.5}{95} \times [1 + 0.5 (3 - 1)] = 1$

Total number of moles = 1.75

Mass of water = $100 - (29.25 + 47.5) = 23.25$

$T_{b} = 100 + [1.75 \times 0.532 \times \frac{1000}{23.25}]$ =140.04

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