MHT CET · Chemistry · Electrochemistry
The resistance of a conductivity cell of 0.1 M KCl solution is 120 ohm and conductivity is \(1.64 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}\). What is the value of cell constant?
- A \(0.0136 \mathrm{~cm}^{-1}\)
- B \(0.0618 \mathrm{~cm}^{-1}\)
- C \(0.0196 \mathrm{~cm}^{-1}\)
- D \(0.0731 \mathrm{~cm}^{-1}\)
Answer & Solution
Correct Answer
(C) \(0.0196 \mathrm{~cm}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \text { Conductivity }(\mathrm{k})=\frac{\text { Cell constant }}{\mathrm{R}} \\
& \therefore \quad \text { Cell constant }=\mathrm{k} \times \mathrm{R} \\
& =1.64 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1} \times 120 \Omega \\
& =0.0196 \mathrm{~cm}^{-1}
\end{aligned}\)
& \text { Conductivity }(\mathrm{k})=\frac{\text { Cell constant }}{\mathrm{R}} \\
& \therefore \quad \text { Cell constant }=\mathrm{k} \times \mathrm{R} \\
& =1.64 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1} \times 120 \Omega \\
& =0.0196 \mathrm{~cm}^{-1}
\end{aligned}\)
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