Sea water contains 29.25% NaCl and 19% by weight of solution. The normal boiling point of the sea water is ______ °C (Nearest integer) Assume 100% ionization for both NaCl and Given: = 0.52 K kg Molar mass of NaCl and is 58.5 and 95 g respectivel

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Solution

Q.
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. [2023]

Ans.
(116)

$Total weight of solute in solution = 29.25 + 19 = 48.25 gm$

$Total weight of solvent in solution = 100 - 48.25 = 51.75 gm$

So $Δ T_{b} = i \times K_{b} \times m = (i \times m) \times K_{b}$

$Mole of NaCl = \frac{29.25}{58.5} = 0.5$

${Mole of MgCl}_{2} = \frac{19}{95} = 0.2$

$Δ T_{b} = (2 \times \frac{29.25}{58.5} \times \frac{1000}{51.75} + 3 \times \frac{19}{95} \times \frac{1000}{51.75}) \times 0.52$

$Δ T_{b} = 16.075$

$Δ T_{b} = {(T_{b})}_{solution} - {(T_{b})}_{_{solvent}}$

${(T_{b})}_{solution} = 100 + 16.07$

$= 116.07 ° C$

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