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The limiting molar conductivities Λ o for NaCl, KBr and KCl are 126, 152 and 150 S c m 2 m o l - 1 respectively. The Λ o for NaBr is

Options

  1. A278 S c m 2 m o l - 1
  2. B178 S c m 2 m o l - 1
  3. C128 S c m 2 m o l - 1
  4. D306 S c m 2 m o l - 1

Correct answer

C. 128 S c m 2 m o l - 1

Step-by-step solution

By kohlrausch’s law Λ o N a B r = Λ o N a C l + Λ o K B r - Λ o K C l = 126 + 152 – 150 = 128 S c m 2 m o l - 1

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