The molar conductivities at infinite dilution of ${NH}_{4} Cl, HCl$ and $HCN$ are 150,426 and $432 S {cm}^{2} {mol}^{- 1}$ then the molar conductivity of ${NH}_{4} CN$ at infinite dilution will be

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$156 S {cm}^{2} {mol}^{- 1}$

$288 S {cm}^{2} {mol}^{- 1}$

$144 S {cm}^{2} {mol}^{- 1}$

$312 S {cm}^{2} {mol}^{- 1}$

answer is A.

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Detailed Solution

$Λ_{m}^{\infty} ({NH}_{4} CN) = Λ_{m}^{\infty} ({NH}_{4} Cl) + Λ_{m}^{\infty} (HCN) - Λ_{m}^{\infty} (HCl) = (150 + 432 - 426) S {cm}^{2} {mol}^{- 1} = 156 S {cm}^{2} {mol}^{- 1}$

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