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Q.

The limiting equivalent conductivity of NaCl, KCI and KBr are 126.5, 150.0 and151.5 S cm²eq^-1, respectively. The limiting equivalent ionic conductance for Br^- is 78 S cm² eq^-1. The limiting equivalent ionic conductance for Na⁺ ions would be ----- S cm² eq^-1

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a

125

b

50

c

128

d

49

answer is D.

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Detailed Solution

$\begin{array}{l} Λ_{eq}^{\infty} (NaBr) = Λ_{eq}^{\infty} (NaCl) + Λ_{eq}^{\infty} KBr - Λ_{eq}^{\infty} (KCl) \\ λ_{eq}^{\infty} ({Na}^{+}) + λ_{eq}^{\infty} ({Br}^{-}) = 126 .5 + 151.5 - 150 \\ λ_{eq}^{\infty} ({Na}^{+}) = 128 - 78 = 50 S {cm}^{2} {eq}^{- 1} \end{array}$

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