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

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$\Lambda_m^{\large\circ}$ 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.

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