MHT CET · Chemistry · Electrochemistry
What is the cell constant, if two platinum electrodes in conductivity cell are separated by 0.92 cm and area of cross section is \(1.2 \mathrm{~cm}^2\) ?
- A \(0.767 \mathrm{~cm}^{-1}\)
- B \(0.906 \mathrm{~cm}^{-1}\)
- C \(1.304 \mathrm{~cm}^{-1}\)
- D \(1.104 \mathrm{~cm}^{-1}\)
Answer & Solution
Correct Answer
(A) \(0.767 \mathrm{~cm}^{-1}\)
Step-by-step Solution
Detailed explanation
Cell constant \(=\frac{l}{\mathrm{a}}=\frac{0.92 \mathrm{~cm}}{1.2 \mathrm{~cm}^2}=0.767 \mathrm{~cm}^{-1}\)
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