Q33 / 120 · VITEEE 2007 · Chemistry

If the molar conductance values of $\mathrm{Ca}^{2+}$ and $\mathrm{Cl}^{-}$ at infinite dilution are respectively $118.88 \times 10^{-4}$ $\mathrm{m}^{2} \mathrm{mh} \mathrm{mol}^{-1}$ and $77.33 \times 10^{-4} \mathrm{~m}^{2} \mathrm{mho} \mathrm{mol}^{-1}$ then that of $\mathrm{CaCl}_{2}$ is (in $\mathrm{m}^{2} \mathrm{mhomol}^{-1}$)

  • A
    $118.88 \times 10^{-4}$
  • B
    $154.66 \times 10^{-4}$
  • Correct answer
    C
    $273.54 \times 10^{-4}$
  • D
    $196.21 \times 10^{-4}$

Solution

$\begin{aligned} \text { Molar conductance of } \mathrm{CaCl}_{2} & \\ &=\text { Molar conductance of } \mathrm{Ca}^{2++}+2 \times(\text { molar }\\ &\left.\text { conductance of } \mathrm{Cl}^{-}\right) \\ &=118.88 \times 10^{-4}+2\left(77.33 \times 10^{-4}\right) \\ &=273.54 \times 10^{-4} \mathrm{~m}^{2} \text { mhomol }^{-1} \end{aligned}$

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

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