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NEET CHEMISTRYElectrochemistryMedium

Question

The molar conductance of M/32 solution of a weak monobasic acid is 8.0 ohm1 cm28.0 \text{ ohm}^{-1} \text{ cm}^2 and at infinite dilution is 400 ohm1 cm2400 \text{ ohm}^{-1} \text{ cm}^2. The dissociation constant of this acid is:

A

1.25×1051.25 \times 10^{-5}

B

1.25×1061.25 \times 10^{-6}

C

6.25×1046.25 \times 10^{-4}

D

1.25×1041.25 \times 10^{-4}

Step-by-Step Solution

  1. Calculate Degree of Dissociation (α\alpha): For a weak electrolyte, the degree of dissociation α\alpha is the ratio of molar conductivity at a specific concentration (Λm\Lambda_m) to the limiting molar conductivity (Λm\Lambda_m^\circ). α=ΛmΛm\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} Given: Λm=8.0 S cm2 mol1\Lambda_m = 8.0 \text{ S cm}^2 \text{ mol}^{-1} and Λm=400 S cm2 mol1\Lambda_m^\circ = 400 \text{ S cm}^2 \text{ mol}^{-1}. α=8.0400=0.02\alpha = \frac{8.0}{400} = 0.02

  2. Calculate Dissociation Constant (KaK_a): The dissociation constant for a weak acid is given by: Ka=cα21αK_a = \frac{c\alpha^2}{1 - \alpha} Given: Concentration c=M/32=1/32 mol L10.03125 Mc = \text{M}/32 = 1/32 \text{ mol L}^{-1} \approx 0.03125 \text{ M}. Since α(0.02)\alpha (0.02) is very small compared to 1, we can approximate 1α11 - \alpha \approx 1. Kacα2K_a \approx c\alpha^2 Ka=132×(0.02)2K_a = \frac{1}{32} \times (0.02)^2 Ka=132×4×104K_a = \frac{1}{32} \times 4 \times 10^{-4} Ka=18×104=0.125×104K_a = \frac{1}{8} \times 10^{-4} = 0.125 \times 10^{-4} Ka=1.25×105K_a = 1.25 \times 10^{-5}

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Electrochemistry. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYElectrochemistryconductancesolutionmonobasicinfinitedilution

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