# From following molar conductivities at infinite dilution $\Lambda_m^{\large\circ}$ for $Ba(OH)_2=457.6\Omega^{-1}cm^2mol^{-1}$,$\Lambda_m^{\large\circ}$ for $BaCl_2=240.6\Omega^{-1}cm^2mol^{-1}$,$\Lambda_m^{\large\circ}$ for $NH_4Cl=129.8\Omega^{-1}cm^2mol^{-1}$.The $\Lambda_m^{\large\circ}$ for $NH_4OH$ is

$\begin{array}{1 1}(a)\;238.3\Omega^{-1}cm^2mol^{-1}&(b)\;262.3\Omega^{-1}cm^2mol^{-1}\\(c)\;289\Omega^{-1}cm^2mol^{-1}&(d)\;242\Omega^{-1}cm^2mol^{-1}\end{array}$

$\Lambda_m^{\large\circ}$ for $Ba(OH)_2=\Lambda_{Be^{+2}}^{\large\circ}+2\Lambda_{OH^{-}}^{\large\circ}$-----(1)
$\Lambda_m^{\large\circ}$ for $BaCl_2=\Lambda_{NH_4^{+}}^{\large\circ}+2\Lambda_{Cl^{-}}^{\large\circ}$-----(2)
$\Lambda_m^{\large\circ}$ for $(NH_4Cl)=\Lambda_{NH_4^{+}}^{\large\circ}+\Lambda_{Cl^{-}}^{\large\circ}$-----(3)
$\Lambda_m^{\large\circ}$ = $(NH_4OH)=\Lambda ^{\large\circ}_{NH_4^+}+\Lambda ^{\large\circ}_{OH^-}$
$\Rightarrow \large\frac{1}{2}$$equ(1)+equ(3)-\large\frac{1}{2}$$equ(2)$
$\Rightarrow \large\frac{1}{2}$$\times 457.6+129.8-\large\frac{1}{2}$$240.6$
$\Rightarrow 238.3\Omega^{-1}cm^2mol^{-1}$
Hence (a) is the correct answer.