COMEDK · Chemistry · 18. Electrochemistry
\(0.1 \mathrm{M}\) solution of \(\mathrm{AgNO}_3\) is taken in a Conductivity cell and a potential difference of \(40 \mathrm{~V}\) is applied across the ends of a column of this solution whose diameter is \(4.0 \mathrm{~cm}\) and length of the column is \(12 \mathrm{~cm}\). The current used is \(0.4 \mathrm{~A}\). The Molar conductivity of the solution is _________.
- A \(0.009546 \mathrm{~Scm}^2 \mathrm{~mol}^{-1}\)
- B \(954.7 \mathrm{~Scm}^2 \mathrm{~mol}^{-1}\)
- C \(95.5 \mathrm{~Scm}^2 \mathrm{~mol}^{-1}\)
- D \(9.547 \mathrm{~Scm}^2 \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(C) \(95.5 \mathrm{~Scm}^2 \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
The resistance \(R\) of the solution is given by Ohm's law: \(R = \dfrac{V}{I} = \dfrac{40 \text{ V}}{0.4 \text{ A}} = 100 \text{ } \Omega\). The area of cross-section \(A\) of the column is \(\pi r^2\). Given diameter \(d = 4.0 \text{ cm}\), the radius \(r = 2.0 \text{ cm}\).…
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