Question
From the following molar conductivities at infinite dilution, calculate $\Lambda^0_{\text{m}}$ for $NH_4OH$.
$\Lambda^0_{\text{m}}\text{ for }\text{Ba(OH)}_2=457.6\Omega^{-1}\text{cm}^2\text{mol}^{-1}$
$\Lambda^0_{\text{m}}\text{ for }\text{BaCl}_2=240.\Omega^{-1}\text{cm}^2\text{mol}^{-1}$
$\Lambda^{0}_{\text{m}}\text{ for }\text{NH}_4\text{Cl}=129.8\Omega^{-1}\text{cm}^2\text{mol}^{-1}$

Answer

$\Lambda^0_{\text{m}(\text{NH}_4\text{OH})}=\lambda^0_{\text{NH}^+_4}+\lambda^{0}_{\text{OH}^-}$
$=\Big(\lambda^0_{\text{NH}^+_4}+\lambda^0_{\text{Cl}^{-}}\Big)+\frac{1}{2}\Big(\lambda^0_{\text{Ba}^{2+}}+2\lambda^0_{\text{OH}^-}\Big)\\-\frac{1}{2}\Big(\lambda^0_{\text{Ba}^{2+}}+2\lambda^0_{\text{Cl}^-}\Big)$
$\Lambda^0_{\text{m}(\text{NH}_4\text{Cl})}+\frac{1}{2}\Big[\Lambda^0_{\text{m}(\text{BaOH})_2}\Big]-\frac{1}{2}\Big[\Lambda^0_{\text{m(BaCl}_2)}\Big]$
$129.8+\frac{1}{2}\times457.6-\frac{1}2{}\times240.6$
$=238.3\text{ ohm}^{-1}\text{cm}^2\text{mol}^{-1}$

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