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?
- A \(0.616 \mathrm{~cm}^{-1}\)
- B \(0 \cdot 491 \mathrm{~cm}^{-1}\)
- C \(2 \cdot 0 \mathrm{~cm}^{-1}\)
- D \(0 \cdot 2 \mathrm{~cm}^{-1}\)
Answer & Solution
Correct Answer
(A) \(0.616 \mathrm{~cm}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{k}=0.0112 \Omega^{-1} \mathrm{~cm}^{-1}, \mathrm{R}=55.0 \Omega\)
\(\text {Cell constant, } \mathrm{b} =\mathrm{k} \times \mathrm{R}\)
\(=0.0112 \Omega^{-1} \mathrm{~cm}^{-1} \times 55.0 \Omega\)
\(\therefore \mathrm{b} =0.616 \mathrm{~cm}^{-1}\)
\(\text {Cell constant, } \mathrm{b} =\mathrm{k} \times \mathrm{R}\)
\(=0.0112 \Omega^{-1} \mathrm{~cm}^{-1} \times 55.0 \Omega\)
\(\therefore \mathrm{b} =0.616 \mathrm{~cm}^{-1}\)
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