Start Learning Free

Before You Begin

Question #44745

Difficulty: 8/10 | 🟢 Free

The Question

Consider a 0.05 M0.05 \text{ M} aqueous solution of a weak electrolyte \ce{AB2} (theoretical molar mass = 300 g mol−1300 \text{ g mol}^{-1}) that partially dissociates into \ce{A^{2+}} and \ce{B^-} ions at 298 K298 \text{ K}. The electrolytic conductivity (κ\kappa) of this solution is 1.2×10−3 S cm−11.2 \times 10^{-3} \text{ S cm}^{-1}.

The limiting molar conductivities (Λm∞\Lambda_m^\infty) of various strong electrolytes at 298 K298 \text{ K} are given as: Λm∞(\ceACl2)=260 S cm2 mol−1\Lambda_m^\infty(\ce{ACl2}) = 260 \text{ S cm}^2 \text{ mol}^{-1} Λm∞(\ceNaB)=110 S cm2 mol−1\Lambda_m^\infty(\ce{NaB}) = 110 \text{ S cm}^2 \text{ mol}^{-1} Λm∞(\ceNaCl)=120 S cm2 mol−1\Lambda_m^\infty(\ce{NaCl}) = 120 \text{ S cm}^2 \text{ mol}^{-1}

Based on the interplay between electrolytic conductance and colligative properties, what is the observed molar mass of \ce{AB2} in this solution?

Options

  • A.

    250 g mol−1250 \text{ g mol}^{-1}

  • B.

    100 g mol−1100 \text{ g mol}^{-1}

  • C.

    272.7 g mol−1272.7 \text{ g mol}^{-1}

  • D.

    251.7 g mol−1251.7 \text{ g mol}^{-1}

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

Degree of Dissociation (Alpha)|Molecular Mass Abnormalities|Resistance and Resistivity Interconversion

Why This Question Is Tricky

Common Mistakes

This problem tests the critical bridge between Electrochemistry and Solution Chemistry.

The primary trap lies in the calculation of the Van 't Hoff factor (ii).

Students frequently assume complete dissociation (i=3i=3) for salts, ignoring the weak electrolyte nature indicated by the conductance data, which leads to an incorrect molar mass of 100 g mol−1100 \text{ g mol}^{-1}.

A secondary trap involves failing to account for the stoichiometry in Kohlrausch's Law (forgetting to multiply \ce{NaB} and \ce{NaCl} by 2), which skews Λm∞\Lambda_m^\infty, α\alpha, and ultimately MobsM_{obs} (leading to 251.7 g mol−1251.7 \text{ g mol}^{-1}).

Finally, students must correctly apply the inverse relationship between the Van 't Hoff factor and observed molar mass (Mobs=Mtheo/iM_{obs} = M_{theo}/i).

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterElectrochemistry - Class 12
ConceptDegree of Dissociation (Alpha)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree