Molar conductivities at infinite dilution $\wedge_{\mathrm{m}}^{\circ}$ for $\mathrm{Ba}(\mathrm{OH})_2$,…

Molar conductivities at infinite dilution $\wedge_{\mathrm{m}}^{\circ}$ for $\mathrm{Ba}(\mathrm{OH})_2$, $\mathrm{BaCl}_2$ and $\mathrm{NH}_4 \mathrm{Cl}$ are $457.0,240.6$ and $2130 \mathrm{Scm}^2 \mathrm{~mol}^{-1}$ respectively. The $\wedge_{\mathrm{m}}^0$ for ammonium hydroxide (in $\mathrm{Scm}^2$ $\mathrm{mol}^{-1}$ ) is
  1. 1683.2
  2. 1080.2
  3. 2130.0
  4. 2238.2

Solution

$\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{NH}_4 \mathrm{OH}\right)=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{NH}_4^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{OH}^{-}\right)$ $\begin{aligned} & =\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{NH}_4 \mathrm{Cl}\right)+\frac{1}{2} \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Ba}(\mathrm{OH})_2\right)-\frac{1}{2} \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{BaCl}_2\right) \\ & =2130+\left(\frac{1}{2} \times 457.0\right)-\left(\frac{1}{2} \times 240.6\right) \\ & =2238.2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{aligned}$

Asked in: AP EAMCET 2023 (16 May Shift 1)

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