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MHT CET · Chemistry · Electrochemistry

What is the cell constant of \(\frac{\mathrm{N}}{10} \mathrm{KCl}\) solution at \(25^{\circ} \mathrm{C}\), if conductivity and resistance of a solution is \(0.0112 \Omega^{-1} \mathrm{~cm}^{-1}\) and \(55 \cdot 0 \Omega\) respectively?

  1. A \(0.616 \mathrm{~cm}^{-1}\)
  2. B \(0 \cdot 491 \mathrm{~cm}^{-1}\)
  3. C \(2 \cdot 0 \mathrm{~cm}^{-1}\)
  4. D \(0 \cdot 2 \mathrm{~cm}^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(0.616 \mathrm{~cm}^{-1}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{k}=0.0112 \Omega^{-1} \mathrm{~cm}^{-1}, \mathrm{R}=55.0 \Omega\)
\(\text {Cell constant, } \mathrm{b} =\mathrm{k} \times \mathrm{R}\)
\(=0.0112 \Omega^{-1} \mathrm{~cm}^{-1} \times 55.0 \Omega\)
\(\therefore \mathrm{b} =0.616 \mathrm{~cm}^{-1}\)