MHT CET · Chemistry · Electrochemistry
What is the conductivity of \(0.02 \mathrm{M} \mathrm{KCl}\) solution if cell constant is \(1.29 \mathrm{~cm}^{-1}\) with resistance \(645 \Omega\) ?
- A \(5.0 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
- B \(2.0 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
- C \(8.3 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
- D \(2.5 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
Answer & Solution
Correct Answer
(B) \(2.0 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{K}=\frac{1}{\mathrm{R}} \cdot \frac{\ell}{\mathrm{A}} \quad\left(\text { Cell constant }=\frac{\ell}{\mathrm{A}}\right) \\ & \mathrm{K}=\frac{1}{645} \times 1.29 \\ & =0.002 \\ & =2 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\end{aligned}\)
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