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Question #44527

Difficulty: 8/10 | 🟢 Free

The Question

An aqueous solution of \ce{K4[Fe(CN)6]} has a molality of $0.40 \text{ mol kg}^{-1}$ . To purify the water via reverse osmosis at $300 \text{ K}$ , a minimum applied pressure of $39.28 \text{ bar}$ is required to overcome the osmotic pressure. Assume the solution follows the exact thermodynamic relation for real solutions: $\Pi = -\frac{RT}{V_m} \ln(x_{\text{solvent}})$ , where $V_m$ is the molar volume of pure water ( $0.018 \text{ L mol}^{-1}$ ). Given: $R = 0.083 \text{ L bar K}^{-1} \text{ mol}^{-1}$ , $e^{-0.0284} = 0.9720$ , and moles of water in $1 \text{ kg} = 55.55 \text{ mol}$ . Calculate the percentage degree of dissociation ( $\alpha$ ) of the complex in this solution.

Options

A.
$75.0\%$
B.
$73.6\%$
C.
$80.0\%$
D.
$100\%$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Vapor Pressure Lowering (Mole Fraction)|Reverse Osmosis Dynamics|Stoichiometric Coefficient Division

Why This Question Is Tricky

Common Mistakes

This problem tests the transition from ideal dilute approximations to real solution thermodynamics in Reverse Osmosis.

Students commonly fall into the trap of using the linear Van't Hoff equation ( $\Pi = i m RT$ ), which incorrectly yields $73.6\%$ .

Additionally, failing to correctly identify the stoichiometric coefficient of the complex ion ( $n=5$ ) or misapplying the dissociation formula ( $i = n\alpha$ ) leads to the distractors $100\%$ and $80.0\%$ .

The core exception is that at high pressures, activity-based mole fractions must be used instead of linear molarity approximations.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Solutions - Class 12
Concept	Vapor Pressure Lowering (Mole Fraction)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free