Exactly of mercuric nitrate, (Molar mass = ), is dissolved in of water. The freezing point is measur — 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 dissolved in $1000\text{ g}$ of water. The freezing point is measured to be $-0.0558^\circ\text{C}$. Given $K_f = 1.86\text{ K kg mol}^{-1}$, calculate the percentage degree of ionization ($\% \alpha$).
Answer: C
💡 Solution & Explanation
Molality $m = \frac{3.24/324}{1} = 0.01\text{ m}$. $\Delta T_f = i \times K_f \times m \implies 0.0558 = i \times 1.86 \times 0.01 \implies i = 3$. The salt dissociates as $Hg^{2+} + 2NO_3^-$, giving $n=3$. Since $i = n$, the degree of ionization $\alpha = 1$, which is exactly $100\%$.
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