Q13 / 50 · Manipal (MET) 2022 · Chemistry
From the following molar conductivities at infinite dilution, $\Lambda_{\mathrm{m}}^{\circ}$ (in $\Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ ) for $\mathrm{NH}_4 \mathrm{OH}$ is
$\qquad$
$\begin{aligned} & \Lambda_{\mathrm{m}}^{\circ} \text { for } \mathrm{Ba}(\mathrm{OH})_2=446.8 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ & \Lambda_{\mathrm{m}}^{\circ} \text { for } \mathrm{BaCl}_2=241.6 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ & \Lambda_{\mathrm{m}}^{\circ} \text { for } \mathrm{NH}_4 \mathrm{Cl}=130 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{aligned}$
Correct answer: 232.6
Solution
$\begin{aligned} & \Lambda_{\mathrm{m}\left(\mathrm{NH} \mathrm{H}_4 \mathrm{OH}\right)=}^{\circ}= \Lambda_{\mathrm{NH}_4^{+}}^{\circ}+ \\ &= \Lambda_{\mathrm{OH}^{-}}^{\circ} \\ &=\left(\Lambda_{\mathrm{NH}_4^{+}}^{\circ}+\Lambda_{\mathrm{Cl}^{-}}^{\circ}\right)+\frac{1}{2}\left(\Lambda_{\mathrm{Ba}^{2+}}^{\circ}+2 \Lambda_{\mathrm{OH}^{-}}^{\circ}\right) \\ &-\frac{1}{2}\left(\Lambda_{\mathrm{Ba}^{2+}}^{\circ}+2 \Lambda_{\mathrm{Cl}^{-}}^{\circ}\right) \\ &= \Lambda_{\mathrm{m}\left(\mathrm{NH}_4 \mathrm{Cl}\right)}^{\circ}+\frac{1}{2} \Lambda_{\mathrm{m}\left[\mathrm{Ba}(\mathrm{OH})_2\right]}^{\circ}-\frac{1}{2} \Lambda_{\mathrm{m}\left(\mathrm{BaCl}_2\right)}^{\circ} \\ &= 130+\frac{1}{2}+446.8-\frac{1}{2} \times 241.6 \\ &= 232.6 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1} .\end{aligned}$
Hence, the correct answer is 232.6.