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NEET 2026
Exam
NEET
Year
2026
Subject
Chemistry
Chapter
Chapter

NEET 2026 · Chemistry · Chapter

The molar conductivity of a 0.038 M solution of a weak base MOH is...

The molar conductivity of a 0.038 M0.038 \text{ M} solution of a weak base MOHMOH is 10 S cm2 mol−110 \text{ S cm}^2 \text{ mol}^{-1}. What is the dissociation constant (KbK_b) of the base?

Given:
λM+∘=50 S cm2 mol−1\lambda^\circ_{M^+} = 50 \text{ S cm}^2 \text{ mol}^{-1}
λOH−∘=150 S cm2 mol−1\lambda^\circ_{OH^-} = 150 \text{ S cm}^2 \text{ mol}^{-1}

Options

  1. A9.5×10−5 mol L−19.5 \times 10^{-5} \text{ mol L}^{-1}
  2. B1.0×10−4 mol L−11.0 \times 10^{-4} \text{ mol L}^{-1}
  3. C2.0×10−3 mol L−12.0 \times 10^{-3} \text{ mol L}^{-1}
  4. D1.0×10−5 mol L−11.0 \times 10^{-5} \text{ mol L}^{-1}

Correct answer

B. 1.0×10−4 mol L−11.0 \times 10^{-4} \text{ mol L}^{-1}

Step-by-step solution

The limiting molar conductivity of MOHMOH is given by Kohlrausch's law:

Λm∘(MOH)=λM+∘+λOH−∘\Lambda^\circ_m(MOH) = \lambda^\circ_{M^+} + \lambda^\circ_{OH^-}

Λm∘(MOH)=50+150=200 S cm2 mol−1\Lambda^\circ_m(MOH) = 50 + 150 = 200 \text{ S cm}^2 \text{ mol}^{-1}

The degree of dissociation α\alpha is:

α=ΛmΛm∘=10200=0.05\alpha = \dfrac{\Lambda_m}{\Lambda^\circ_m} = \dfrac{10}{200} = 0.05

The dissociation constant KbK_b for the weak base is:

Kb=Cα21−αK_b = \dfrac{C\alpha^2}{1-\alpha}

Substituting the values:

Kb=0.038×(0.05)21−0.05K_b = \dfrac{0.038 \times (0.05)^2}{1 - 0.05}

Kb=0.038×0.00250.95K_b = \dfrac{0.038 \times 0.0025}{0.95}

Kb=9.5×10−50.95=1.0×10−4 mol L−1K_b = \dfrac{9.5 \times 10^{-5}}{0.95} = 1.0 \times 10^{-4} \text{ mol L}^{-1}

Answer: 1.0×10−4 mol L−11.0 \times 10^{-4} \text{ mol L}^{-1}
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