Exactly of mercuric nitrate, (Molar mass = ), is completely dissolved in of pure water. The freezing — Solutions and Colligative Properties Chemistry Question
Question
Exactly $3.24\text{ g}$ of mercuric nitrate, $Hg(NO_3)_2$ (Molar mass = $324\text{ g mol}^{-1}$), is completely dissolved in $1\text{ kg}$ of pure water. The freezing point of the solution is found to be $-0.0558^\circ\text{C}$. Given $K_f$ for water is $1.86\text{ K kg mol}^{-1}$, calculate the percentage degree of ionization ($\% \alpha$) of the salt.
💡 Solution & Explanation
First, find the molality: $m = \frac{3.24\text{ g} / 324\text{ g mol}^{-1}}{1\text{ kg}} = 0.01\text{ m}$. Using the depression in freezing point formula: $\Delta T_f = i \times K_f \times m \implies 0.0558 = i \times 1.86 \times 0.01 \implies i = \frac{0.0558}{0.0186} = 3$. $Hg(NO_3)_2$ dissociates into 3 ions ($Hg^{2+} + 2NO_3^-$), meaning $n=3$. Using $i = 1 + \alpha(n-1) \implies 3 = 1 + 2\alpha \implies \alpha = 1.0$. The percentage is $100\%$.