A aqueous solution of potassium ferrocyanide, , is supposed to be exactly dissociated at room temper — Solutions and Colligative Properties Chemistry Question
Question
A $1\text{ M}$ aqueous solution of potassium ferrocyanide, $K_4[Fe(CN)_6]$, is supposed to be exactly $40\%$ dissociated at room temperature. Its observed boiling point is found to perfectly match that of another $20\% \text{ (w/v)}$ aqueous solution of a completely non-electrolytic substance $A$. Assuming molarity $\approx$ molality, what is the approximate molecular weight of substance $A$?
💡 Solution & Explanation
$K_4[Fe(CN)_6]$ dissociates into $4 K^+$ and $1 [Fe(CN)_6]^{4-}$, giving $n=5$ ions. At $\alpha = 0.40$, $i = 1 + (5-1) \times 0.40 = 1 + 1.6 = 2.6$. Since the boiling points are identical, their effective concentrations are equal: $\Delta T_b = i_1 C_1 = i_2 C_2$. For the complex: $i \times C = 2.6 \times 1\text{ M} = 2.6$. For substance A ($20\%\text{ w/v}$ means $200\text{ g L}^{-1}$): $i \times C = 1 \times \frac{200}{M_A}$. Equating them: $2.6 = \frac{200}{M_A} \implies M_A = \frac{200}{2.6} \approx 76.92\text{ g mol}^{-1}$, which is closest to $77$.