Question:
A \(0.15\) molal solution of \(\mathrm{K_4[Fe(CN)_6]}\) in water freezes at \(-0.65^\circ\mathrm{C}\). What is the apparent percentage of dissociation of this compound in this solution? (\(K_f\) for water \(= 1.86^\circ\mathrm{C\ mol}^{-1}\))
A \(0.15\) molal solution of \(\mathrm{K_4[Fe(CN)_6]}\) in water freezes at \(-0.65^\circ\mathrm{C}\). What is the apparent percentage of dissociation of this compound in this solution? (\(K_f\) for water \(= 1.86^\circ\mathrm{C\ mol}^{-1}\))
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Use \(i = 1 + (n-1)\alpha\) where \(n\) is the number of ions per formula unit.
Updated On: Apr 23, 2026
- \(0.33\)
- \(0.52\)
- \(0.63\)
- \(0.79\)
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The Correct Option is A
Solution and Explanation
Step 1: Formula / Definition}
\[ \Delta T_f = i \cdot K_f \cdot m \]
Step 2: Calculation / Simplification}
\(\Delta T_f = 0.65\)
\(0.65 = i \times 1.86 \times 0.15 \Rightarrow i = \frac{0.65}{0.279} = 2.33\)
For \(\mathrm{K_4[Fe(CN)_6]}\), \(n = 5\) (dissociates into \(4\mathrm{K}^+ + [\mathrm{Fe(CN)_6}]^{4-}\))
\(i = 1 + (n-1)\alpha \Rightarrow 2.33 = 1 + 4\alpha\)
\(4\alpha = 1.33 \Rightarrow \alpha = 0.3325 \approx 0.33\)
Step 3: Final Answer
\[ 0.33 \]
\[ \Delta T_f = i \cdot K_f \cdot m \]
Step 2: Calculation / Simplification}
\(\Delta T_f = 0.65\)
\(0.65 = i \times 1.86 \times 0.15 \Rightarrow i = \frac{0.65}{0.279} = 2.33\)
For \(\mathrm{K_4[Fe(CN)_6]}\), \(n = 5\) (dissociates into \(4\mathrm{K}^+ + [\mathrm{Fe(CN)_6}]^{4-}\))
\(i = 1 + (n-1)\alpha \Rightarrow 2.33 = 1 + 4\alpha\)
\(4\alpha = 1.33 \Rightarrow \alpha = 0.3325 \approx 0.33\)
Step 3: Final Answer
\[ 0.33 \]
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