MHT CET · Chemistry · Electrochemistry
A conductivity cell containing \(0.001 \mathrm{M} \mathrm{AgNO}_3\) solution develops resistance \(6530 \mathrm{ohm}\) at \(25^{\circ} \mathrm{C}\). Calculate the electrical conductivity of solution at same temperature if the cell constant is \(0.653 \mathrm{~cm}^{-1}\).
- A \(1.3 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}\)
- B \(1.5 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}\)
- C \(1.7 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}\)
- D \(1.0 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}\)
Answer & Solution
Correct Answer
(D) \(1.0 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} & \text { Cell constant }=\mathrm{k} \times \mathrm{R} \\ \therefore & 0.653=\mathrm{k} \times 6530 \\ \therefore & \mathrm{k}=\frac{0.653}{6530}=1 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}\end{array}\)
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