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Limiting molar conductivity of $NH_4OH$ (i.e. $\Lambda_m^\circ(NH_4OH)$) is equal to:
A
$\Lambda_m^\circ(NH_4Cl) + \Lambda_m^\circ(NaOH) - \Lambda_m^\circ(NaCl)$
B
$\Lambda_m^\circ(NH_4Cl) + \Lambda_m^\circ(NaCl) - \Lambda_m^\circ(NaOH)$
C
$\Lambda_m^\circ(NaOH) + \Lambda_m^\circ(NaCl) - \Lambda_m^\circ(NH_4Cl)$
D
$\Lambda_m^\circ(NH_4OH) + \Lambda_m^\circ(NH_4Cl) - \Lambda_m^\circ(HCl)$
Detailed Solution
By Kohlrausch's law, $\Lambda_m^\circ(NH_4OH) = \lambda^\circ(NH_4^+) + \lambda^\circ(OH^-)$
$\Lambda_m^\circ(NH_4Cl) = \lambda^\circ(NH_4^+) + \lambda^\circ(Cl^-)$
$\Lambda_m^\circ(NaOH) = \lambda^\circ(Na^+) + \lambda^\circ(OH^-)$
$\Lambda_m^\circ(NaCl) = \lambda^\circ(Na^+) + \lambda^\circ(Cl^-)$
Adding the first two and subtracting the third cancels $Na^+$ and $Cl^-$:
$\Lambda_m^\circ(NH_4OH) = \Lambda_m^\circ(NH_4Cl) + \Lambda_m^\circ(NaOH) - \Lambda_m^\circ(NaCl)$
$\Lambda_m^\circ(NH_4Cl) = \lambda^\circ(NH_4^+) + \lambda^\circ(Cl^-)$
$\Lambda_m^\circ(NaOH) = \lambda^\circ(Na^+) + \lambda^\circ(OH^-)$
$\Lambda_m^\circ(NaCl) = \lambda^\circ(Na^+) + \lambda^\circ(Cl^-)$
Adding the first two and subtracting the third cancels $Na^+$ and $Cl^-$:
$\Lambda_m^\circ(NH_4OH) = \Lambda_m^\circ(NH_4Cl) + \Lambda_m^\circ(NaOH) - \Lambda_m^\circ(NaCl)$