MHT CET · Chemistry · Electrochemistry
What is the conductivity of 0.05 M KCl solution if cell constant is \(1.32 \mathrm{~cm}^{-1}\) and resistance is 528 ohm?
- A \(0.0401 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\)
- B \(0.0051 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\)
- C \(0.0025 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\)
- D \(0.0691 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\)
Answer & Solution
Correct Answer
(C) \(0.0025 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\text {Conductivity }(\mathrm{k}) =\frac{\text { Cell constant }}{\mathrm{R}} \)
\( =\frac{1.32 \mathrm{~cm}^{-1}}{528 \mathrm{ohm}} \)
\( =0.0025 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\)
\( =\frac{1.32 \mathrm{~cm}^{-1}}{528 \mathrm{ohm}} \)
\( =0.0025 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\)
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