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Electrochemistry question

2004 · Shift 0 · Q38
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Electrochemistry question

2004 · Shift 0 · Q38

JEE MainChemistryElectrochemistryMCQ+4 / −1
The limiting molar conductivities Λ° for NaCl, KBr and KCl are 126, 152 and 150 S cm2 mol-1 respectively. The Λ° for NaBr is
  1. A
    128 S cm2 mol-1
  2. B
    278 S cm2 mol-1
  3. C
    176 S cm2 mol-1
  4. D
    302 S cm2 mol-1
View written solutionFree

Correct answer: A

  1. Use Kohlrausch’s law of independent migration of ions

At infinite dilution, Λ∘(electrolyte)=λ+∘(cation)+λ−∘(anion)\Lambda^\circ(\text{electrolyte}) = \lambda^\circ_+(\text{cation}) + \lambda^\circ_-(\text{anion})Λ∘(electrolyte)=λ+∘​(cation)+λ−∘​(anion)

So, Λ∘(NaCl)=λNa+∘+λCl−∘=126\Lambda^\circ(\text{NaCl}) = \lambda^\circ_{\text{Na}^+} + \lambda^\circ_{\text{Cl}^-} = 126Λ∘(NaCl)=λNa+∘​+λCl−∘​=126 Λ∘(KBr)=λK+∘+λBr−∘=152\Lambda^\circ(\text{KBr}) = \lambda^\circ_{\text{K}^+} + \lambda^\circ_{\text{Br}^-} = 152Λ∘(KBr)=λK+∘​+λBr−∘​=152 Λ∘(KCl)=λK+∘+λCl−∘=150\Lambda^\circ(\text{KCl}) = \lambda^\circ_{\text{K}^+} + \lambda^\circ_{\text{Cl}^-} = 150Λ∘(KCl)=λK+∘​+λCl−∘​=150

We need: Λ∘(NaBr)=λNa+∘+λBr−∘\Lambda^\circ(\text{NaBr}) = \lambda^\circ_{\text{Na}^+} + \lambda^\circ_{\text{Br}^-}Λ∘(NaBr)=λNa+∘​+λBr−∘​

  1. Eliminate the ionic conductivities algebraically

Add the first two equations:

= (\lambda^\circ_{\text{Na}^+} + \lambda^\circ_{\text{Cl}^-}) + (\lambda^\circ_{\text{K}^+} + \lambda^\circ_{\text{Br}^-})$$ Subtract $\Lambda^\circ(\text{KCl}) = \lambda^\circ_{\text{K}^+} + \lambda^\circ_{\text{Cl}^-}$: $$\Lambda^\circ(\text{NaBr}) = \Lambda^\circ(\text{NaCl}) + \Lambda^\circ(\text{KBr}) - \Lambda^\circ(\text{KCl})$$ 3. **Substitute the values** $$\Lambda^\circ(\text{NaBr}) = 126 + 152 - 150 = 128\ \text{S cm}^2\text{ mol}^{-1}$$ 4. **Match with the options** $$128\ \text{S cm}^2\text{ mol}^{-1}$$ corresponds to **Option A**. 5. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** So, the derived answer agrees with the stored answer.
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