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Question #44100

Difficulty: 8/10 | 🟢 Free

The Question

A rigid container of total volume 3.0 L3.0\text{ L} is divided into two compartments, A and B, by a freely movable, strictly solvent-permeable semi-permeable membrane. Compartment A initially has a volume of 2.0 L2.0\text{ L} and contains an aqueous solution comprising 0.20 mol0.20\text{ mol} of \ceMNO3\ce{MNO3} and 0.20 mol0.20\text{ mol} of \ceNaX\ce{NaX}. Compartment B initially has a volume of 1.0 L1.0\text{ L} and contains an aqueous solution of 1.20 mol1.20\text{ mol} of sucrose (a non-electrolyte). For the sparingly soluble salt \ceMX\ce{MX}, the solubility product is Ksp=1.0×10−2K_{sp} = 1.0 \times 10^{-2} at 298 K298\text{ K}. Assume ideal behavior for all solutions, complete dissociation of strong electrolytes, and that the volume of the solid precipitate is negligible. When the system reaches thermodynamic equilibrium at 298 K298\text{ K}, what is the mass of \ceMX\ce{MX} precipitated? (Given: Molar mass of \ceMX=150 g mol−1\ce{MX} = 150\text{ g mol}^{-1})

Options

  • A.

    12.0 g

  • B.

    15.0 g

  • C.

    18.0 g

  • D.

    30.0 g

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

Solubility Product (Ksp) vs. Ionic Product (Qsp)|Isotonic, Hypertonic, Hypotonic

Why This Question Is Tricky

Common Mistakes

This problem tests the coupling of colligative properties with ionic equilibrium.

The 'Hidden Variable' is the dynamic loss of osmolarity due to precipitation.

Students commonly fall into the trap of calculating the final volume based solely on the initial moles of solute (setting 0.80/V=1.20/(3.0−V)0.80/V = 1.20/(3.0-V) to get V=1.20 LV = 1.20\text{ L}), completely forgetting that as \ceMX\ce{MX} precipitates, it removes osmotically active particles from the solution.

This shifts the osmotic equilibrium further, leading to a coupled state.

Ignoring this dynamic loss yields y=0.08 moly = 0.08\text{ mol}, leading to the trap answer of 12.0 g.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterEquilibrium - Class 11 Part 1
ConceptSolubility Product (Ksp) vs. Ionic Product (Qsp)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree