A strong electrolyte with the formula (normal molar mass = ) is dissolved in water. The observed mol — Solutions and Colligative Properties Chemistry Question
Question
A strong electrolyte with the formula $AB_2$ (normal molar mass = $164\text{ g mol}^{-1}$) is dissolved in water. The observed molar mass of the solute in the solution is experimentally determined by colligative methods to be $65.6\text{ g mol}^{-1}$. Calculate the precise percentage degree of dissociation ($\% \alpha$) of the electrolyte.
Answer: 75
💡 Solution & Explanation
The van't Hoff factor is the ratio of normal to observed molar mass: $i = \frac{M_{normal}}{M_{observed}} = \frac{164}{65.6} = 2.5$. The electrolyte $AB_2$ dissociates into 3 ions ($A^{2+} + 2B^-$), so $n = 3$. The relation is $i = 1 + \alpha(n-1) \implies 2.5 = 1 + \alpha(3-1) \implies 1.5 = 2\alpha \implies \alpha = 0.75$. Converting to percentage gives $75\%$.
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