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