When exactly of an unknown binary electrolyte (molecular weight = ) is completely dissolved in of pu — Solutions and Colligative Properties Chemistry Question
Question
When exactly $10\text{ g}$ of an unknown binary electrolyte (molecular weight = $100\text{ g mol}^{-1}$) is completely dissolved in $500\text{ g}$ of pure water, the freezing point of the solution is observed to be $-0.74^\circ\text{C}$. Given $K_f = 1.85\text{ K kg mol}^{-1}$ for water, what is the degree of ionization ($\alpha$) of the electrolyte?
Answer: C
💡 Solution & Explanation
Molality $m = \frac{10}{100} \times \frac{1000}{500} = 0.2\text{ m}$. The depression in freezing point is $\Delta T_f = i \times K_f \times m \implies 0.74 = i \times 1.85 \times 0.2 \implies 0.74 = i \times 0.37 \implies i = 2$. For a binary electrolyte, $n=2$. Since $i = 1 + \alpha(n-1) \implies 2 = 1 + \alpha(2-1) \implies \alpha = 1$, which is exactly $100\%$.
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