Verified Solution • 2025–26 Session moderate

QNEET-2021-Ch-585: NEET 2021 Chemistry - Class 12 NEET-2021-Ch-585 Solution: The molar conductivity of 0.007 M acetic acid is \(20\ {S…

The molar conductivity of 0.007 M acetic acid is 20 S cm2 mol−120\ \text{S cm}^2\ mol^{-1}. What is the dissociation constant of acetic acid? Choose the correct option. [ΛH+∘=350 S cm2 mol−1\Lambda_{H^+}^\circ = 350\ \text{S cm}^2\ mol^{-1} ΛCH3COO−∘=50 S cm2 mol−1\Lambda_{CH_3COO^-}^\circ = 50\ \text{S cm}^2\ mol^{-1}] 1.75×10−5 mol L−11.75 \times 10^{-5}\ \text{mol L}^{-1} 2.50×10−5 mol L−12.50 \times 10^{-5}\ \text{mol L}^{-1} 1.75×10−4 mol L−11.75 \times 10^{-4}\ \text{mol L}^{-1} 2.50×10−4 mol L−12.50 \times 10^{-4}\ \text{mol L}^{-1}

Multiple Choice Options

  • A 1.75 ×\times _ 10 −- _ 5 mol L −- _ 1
  • B 2.50 ×\times _ 10 −- _ 5 mol L −- _ 1
  • C 1.75 ×\times _ 10 −- _ 4 mol L −- _ 1
  • D 2.50 ×\times _ 10 −- _ 4 mol L −- _ 1

Step-by-Step Verified Solution

_ m = 20 S cm 2 mol _ 1 According to Kohlrausch�s law, _ m C H 3 C O O H o = _ C H 3 C O O _ o + _ m H + o = 50 + 350 = 400 S cm 2 mol _ 1 Degree of dissociation, _ = _ m _ m o = 20 400 = 1 20 K a = C _ 2 1 _ _ _ C _ 2 As _ << 1 so 1 - _ � 1 = 7 _ 10 _ 3 _ ( 1 20 ) 2 = 7 _ 10 _ 3 _ 1 4 _ 10 _ 2 = 1.75 _ 10 _ 5 mol L -1

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