$Λ_{m}^{0}$ for NH₄Cl, NaOH and NaCl are 130,248 and 126.5 ohm^-1 cm² mol^-1 respectively. The $Λ_{m}^{0}$ of NH₄OH will be

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251.5

244.5

130

504.5

answer is A.

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Detailed Solution

$\land_{m}^{0} [{NH}_{4} OH] = \land_{m}^{0} [{NH}_{4} Cℓ] - \land_{m}^{0} [NaCl] + \land_{m}^{0} [{NH}_{4} OH]$
= 130 + 248 – 126.5
= 251.5 ohm^-1 cm² mol^-1

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