$Λ_{m ({NH}_{4} OH)}^{\circ}$ is equal to .........

Moderate

A

$Λ_{m}^{\circ} ({NH}_{4} OH) + Λ_{m ({NH}_{4} Cl)}^{\circ} - Λ_{m (HCl)}^{\circ}$
B

$Λ_{m ({NH}_{4} Cl)}^{\circ} + Λ_{m}^{\circ} (NaOH)^{\circ} - Λ_{m (NaCl)}^{\circ}$
C

$Λ_{m ({NH}_{4} Cl)}^{\circ} + Λ_{m}^{\circ} (NaCl) - Λ_{m (NaOH)}^{\circ}$
D

$Λ_{m}^{\circ} (NaOH) + Λ_{m}^{\circ} (NaCl) - Λ_{m ({NH}_{4} Cl)}^{\circ}$

Solution

$\begin{aligned} Λ_{m ({NH}_{4} Cl)}^{\circ} = Λ_{m ({NH}_{4}^{+})}^{\circ} + {Λ_{m ({Cl}^{-})}}^{\circ} \\ Λ_{m (NaOH)}^{\circ} = Λ_{m ({Na}^{+})}^{\circ} + Λ_{m ({OH}^{-})} \end{aligned}$
$Λ_{m (NaCl)}^{\circ} = Λ_{m ({Na}^{+})}^{\circ} + Λ_{m ({Cl}^{-})}^{\circ}$
$Λ_{m ({NH}_{4} Cl)}^{\circ} + Λ_{m (NaOH)}^{\circ} - Λ_{m (NaCl)}^{\circ} = Λ_{m ({NH}_{4} OH)}$

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