MHT CET · Chemistry · Electrochemistry
Cell constant of a conductivity cell is \(0.9 \mathrm{~cm}^{-1}\) and resistance shown by \(\mathrm{AgNO}_3\) solution is 6530 ohm. What is the conductivity of \(\mathrm{AgNO}_3\) solution?
- A \(725 \Omega^{-1} \mathrm{~cm}^{-1}\)
- B \(5870 \Omega^{-1} \mathrm{~cm}^{-1}\)
- C \(1.38 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}\)
- D \(4.72 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}\)
Answer & Solution
Correct Answer
(C) \(1.38 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\text {Conductivity }(\mathrm{k}) =\frac{\text { Cell constant }}{\mathrm{R}}=\) \(\frac{0.9 \mathrm{~cm}^{-1}}{6530 \Omega} \)
\( =1.38 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}\)
\( =1.38 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}\)
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