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Abnormal Colligative Properties

2.5 g of a mo...

Question

2.5 g of a monobasic acid when dissolved in 100 g of water elevates the boiling point of the solution by ${0.256}^{\circ} C$ . If 0.5 g of the acid requires 8 milliequivalents of NaOH for complete neutralisation, then the degree of dissociation of the acid is: [ $K_{b}$ of water $= 0.512 K k g m o l^{- 1}$ ]

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Solution

To find molar mass of acid we can simply relate gm equivalents of the acid with the base,
$\frac{0.5}{M} \times 1 = 8 \times 10^{- 3}$
$M = 62.5 gm$
$Δ T_{b} = K_{b} i m \Rightarrow 0.256 = 0.512 (1 + α) m$
$0.256 = 0.512 (1 + α) \times \frac{2.5 \times 1000}{62.5 \times 100}$
$α = 0.25$

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