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The values of equivalent conductivity at infinite dilutions for NH₄Cl, NaOH and NaCl are respectively 149.74, 248.1 and 126.4 ohm^–1 cm² eq^–1. The value of $λ_{eq}^{\infty}$ of NH₄OH is ------------ohm^-1 cm² eq^-1

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71.44

It cannot be calculated from the data given

371.44

271.44

answer is B.

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Detailed Solution

\lambda _{eq\left( {N{H_4}OH} \right)}^o = \lambda _{eq\left( {N{H_4}Cl} \right)}^o + \lambda _{eq\left( {NaOH} \right)}^o - \lambda _{eq\left( {NaCl} \right)}^o

= 149.74 + 248.1 - 126.4 = 271.44Sc{m^2}e{q^{ - 1}}

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