Q33 / 160 · AP EAMCET 2009 · Chemistry

At $25^{\circ} \mathrm{C}$, the molar conductances at infinite dilution for the strong electrolytes $\mathrm{NaOH}, \mathrm{NaCl}$ and $\mathrm{BaCl}_2$ are $248 \times 10^{-4}, 126 \times 10^{-4}$ and $280 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1}$ respectively, $\lambda_m^{\circ} \mathrm{Ba}(\mathrm{OH})_2$ in $\mathrm{Sm}^2 \mathrm{~mol}^{-1}$ is

  • A
    $52.4 \times 10^{-4}$
  • Correct answer
    B
    $524 \times 10^{-4}$
  • C
    $402 \times 10^{-4}$
  • D
    $262 \times 10^{-4}$

Solution

$\begin{aligned} & \mathrm{BaCl}_2+2 \mathrm{NaOH} \longrightarrow \mathrm{Ba}(\mathrm{OH})_2+2 \mathrm{NaCl} \\ & \lambda_{m \mathrm{Ba}(\mathrm{OH})_2}^{\infty}=\lambda_{m \mathrm{BaCl}_2}^{\infty}+2 \lambda_{m \mathrm{NaOH}}^{\infty}-2 \lambda_{m \mathrm{NaCl}}^{\infty}\end{aligned}$ $\begin{aligned} & =280 \times 10^{-4}+2 \times 248 \times 10^{-4} \\ & \quad-2 \times 126 \times 10^{-4} \\ & =(280+496-252) \times 10^{-4} \\ & =524 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1}\end{aligned}$

Hence, the correct answer is option: 'B'.

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