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Colligative Properties

Calculate the...

Question

Calculate the boiling point of solutions when 2 g of $N a_{2} S O_{4} (M = 142 g m o l^{- 1})$ is dissolved in 50 g of water, assuming $N a_{2} S O_{4}$ undergoes complete ionization. ( $K_{b}$ for water = 0.52 K kg $m o l^{- 1}$ )

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Solution

Moles of $N a_{2} S O_{4} = \frac{2 g}{142 g m o l^{- 1}} = 0.014 m o l$

Number of kilograms of solvent $= \frac{50 g}{1000 g k g^{- 1}}$

$= 0.02 k g$

Thus, molality of $N a_{2} S O_{4}$ solution $= \frac{0.014 m o l}{0.02 k g}$

$= 0.7 m o l k g^{- 1}$

For water, change in boiling point,

$△ T_{b} = K_{b} \times m$

$△ T_{b} = 0.52 K k g m o l^{- 1} \times 0.07 m o l k g^{- 1}$

$△ T_{b} = 0.0364 K$

Since water boils at 373.15 K at 1.013 bar pressure, the boiling point of solution will be 373.15 + 0.0364 = 373.1864 K.

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= 0.52

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{100.52}^{o}

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