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Elevation in Boiling Point

Calculate the...

Question

Calculate the boiling point of solution when 4g of $M g S O_{4} (M = 120 g m o l^{- 1})$ was dissolved in 100g of water, assuming $M g S O_{4}$ undergoes complete ionization.
[ $K_{b}$ for water = $0.52 K k g m o l^{- 1}]$

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Solution

The number of moles of $M g S O_{4} n = \frac{m a s s}{molar mass} = \frac{4 g}{120 g / m o l} = 0.03333$ mol.

Mass of water is 100 g or $\frac{100 g}{1000 g / k g} = 0.1$ kg

The molality of the solution is $m = \frac{moles of M g S O_{4}}{mass of water in kg}$

$m = \frac{0.03333 m o l}{0.1 k g} = 0.3333 m$

$M g S O_{4} \to M g^{2 +} + S O_{4}^{2 -}$

The vant Hoff factor $i = 2$ (as dissociation of 1 molecule of magnesium sulphate gives 2 ions)

The elevation in the boiling point $Δ T_{b} = i K_{b} m = 2 \times 0.52 \times 0.3333 = 0.3466 K$

The boiling point of pure water is 373.15 K.

The boiling point of solution will be $373.15 + 0.3466 = 373.4966 K ≃ 373.5 K$ .

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