The observed experimental van't Hoff factor () for a dilute aqueous solution of Barium nitrate, , is — Solutions and Colligative Properties Chemistry Question
Question
The observed experimental van't Hoff factor ($i$) for a dilute $0.1\text{ M}$ aqueous solution of Barium nitrate, $Ba(NO_3)_2$, is carefully measured via osmometry to be exactly $2.74$. Based entirely on this macroscopic observation, what is the precise percentage degree of dissociation ($\% \alpha$) of the salt in this specific aqueous environment?
💡 Solution & Explanation
Barium nitrate dissociates according to $Ba(NO_3)_2 \rightleftharpoons Ba^{2+} + 2NO_3^-$, yielding $n=3$ maximum ions per formula unit. The relationship between the van't Hoff factor and degree of dissociation is $i = 1 + \alpha(n - 1)$. Substituting the given values: $2.74 = 1 + \alpha(3 - 1) \implies 2.74 = 1 + 2\alpha$. Subtracting 1 gives $1.74 = 2\alpha$. Solving for $\alpha$ yields $\alpha = \frac{1.74}{2} = 0.87$. To find the percentage, multiply by 100: $0.87 \times 100 = 87\%$.