COMEDK · Chemistry · 18. Electrochemistry
The specific conductivity of a solution containing \(1.0 \mathrm{~g}\) of anhydrous \(\mathrm{BaCl}_{2}\) in \(200 \mathrm{~cm}^{3}\) of the solution has been found to be \(0.0058 \mathrm{Scm}^{-1}\). The molar and equivalent conductivity of the solution respectively are
- A \(120.83 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\) and \(241.67 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\)
- B \(150.5 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\) and \(289.7 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\)
- C \(241.6 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) and \(120.83 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\)
- D \(248.6 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) and \(180.3 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\)
Answer & Solution
Correct Answer
(C) \(241.6 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) and \(120.83 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\)
Step-by-step Solution
Detailed explanation
The molar mass of \(\mathrm{BaCl}_{2}\) is \(137.3 + 2 \times 35.5 = 208.3 \mathrm{~g/mol}\). The number of moles of \(\mathrm{BaCl}_{2}\) is \(n = \dfrac{1.0 \mathrm{~g}}{208.3 \mathrm{~g/mol}} \approx 0.0048008 \mathrm{~mol}\). The molarity \(M\) of the solution is…
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