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The values of equivalent conductivity at infinite dilutions for ${NH}_{4} Cl, NaOH and NaCl$ are respectively 149.74, 248.1and $126.4 {ohm}^{- 1} {cm}^{- 2} {equi}^{- 1} .$ The value of $λ_{eq}^{\infty} of {NH}_{4} OH$ is

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71.44

It cannot be calculated from the data given

271.44

371.44

answer is B.

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Detailed Solution

${λNH}_{4} Cl + λNaOH - λNaCl$

=149.74+248.1-126.4

=271.44

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