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Question #44722

Difficulty: 8/10 | 🟢 Free

The Question

An aqueous solution is prepared by dissolving WW grams of a weak monobasic acid \ce{HA} (molar mass =120 g mol−1= 120 \text{ g mol}^{-1}) in water to make 500 mL500 \text{ mL} of solution at 298 K298 \text{ K}. This solution is found to be isotonic with a 0.06 M0.06 \text{ M} aqueous solution of glucose at the same temperature. Given the acid dissociation constant KaK_a of \ce{HA} is 2.5×10−32.5 \times 10^{-3} at 298 K298 \text{ K}, determine the value of WW.

Options

  • A.

    3.0 g3.0 \text{ g}

  • B.

    3.6 g3.6 \text{ g}

  • C.

    1.8 g1.8 \text{ g}

  • D.

    4.5 g4.5 \text{ g}

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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

Isotonic, Hypertonic, Hypotonic|Van’t Hoff Factor (i) Application|Degree of Dissociation/Association

Why This Question Is Tricky

Common Mistakes

This problem tests the dependency of the van't Hoff factor (ii) on the degree of dissociation (α\alpha) for weak electrolytes.

A common trap is assuming isotonicity implies equal molarity (C1=C2C_1 = C_2, leading to 3.6 g3.6 \text{ g}) or assuming complete dissociation (i=2i=2, leading to 1.8 g1.8 \text{ g}).

Furthermore, solving the exact equilibrium expression yields a quadratic with two positive roots; students must apply the physical constraint α>0\alpha > 0 (hence C<iCC < iC) to reject the extraneous root (0.075 M0.075 \text{ M}, leading to 4.5 g4.5 \text{ g}).

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterSolutions - Class 12
ConceptIsotonic, Hypertonic, Hypotonic
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree