COMEDK · Chemistry · 18. Electrochemistry
Calculate the molar conductance of \(0.025 \mathrm{M}\) aqueous solution of calcium chloride at \(25^{\circ} \mathrm{C}\). The specific conductance of calcium chloride is \(12.04 \times 10^{-2} \mathrm{Sm}^{-1}\)
- A \(4.816 \times 10^{-5} \mathrm{Sm}^2 \mathrm{~mol}^{-1}\)
- B \(3.816 \times 10^{-5} \mathrm{Sm}^2 \mathrm{~mol}^{-1}\)
- C \(381.6 \times 10^{-5} \mathrm{Sm}^2 \mathrm{~mol}^{-1}\)
- D \(481.6 \times 10^{-5} \mathrm{Sm}^2 \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(D) \(481.6 \times 10^{-5} \mathrm{Sm}^2 \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
The molar conductance \(\Lambda_m\) is given by the formula \(\Lambda_m = \dfrac{\kappa}{C}\), where \(\kappa\) is the specific conductance and \(C\) is the molar concentration. Given: \(\kappa = 12.04 \times 10^{-2} \text{ S m}^{-1}\)…
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