MHT CET · Chemistry · Electrochemistry
The resistance of a conductivity cell containing \(0.001 \mathrm{M}\) KC1 solution at \(300 \mathrm{~K}\) is \(150 \Omega\). What is the cell constant if conductivity of \(\mathrm{KCl}\) solution is \(1.5 \times 10^{-} \Omega^{-1} \mathrm{~cm}^{-1}\)
- A \(0.112 \mathrm{~cm}^{-1}\)
- B \(0.225 \mathrm{~cm}^{-1}\)
- C \(0.337 \mathrm{~cm}^{-1}\)
- D \(0.450 \mathrm{~cm}^{-1}\)
Answer & Solution
Correct Answer
(B) \(0.225 \mathrm{~cm}^{-1}\)
Step-by-step Solution
Detailed explanation
We know \(\mathrm{K}=\frac{1}{\mathrm{R}}=\left(\frac{\mathrm{L}}{\mathrm{A}}\right)\)
\(\begin{aligned} & 1.5 \times 10^{-3}=\frac{1}{150} \times\left(\frac{L}{A}\right) \\ & \frac{L}{A}=0.225 \mathrm{~cm}^{-1}\end{aligned}\)
\(\begin{aligned} & 1.5 \times 10^{-3}=\frac{1}{150} \times\left(\frac{L}{A}\right) \\ & \frac{L}{A}=0.225 \mathrm{~cm}^{-1}\end{aligned}\)
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