MHT CET · Chemistry · Electrochemistry
A conductivity cell containing \(5 \times 10^{-4} \mathrm{M} \mathrm{NaCl}\) solution develops resistance \(14000 \mathrm{ohms}\) at \(25^{\circ} \mathrm{C}\). Calculate the conductivity of solution if the cell constant is \(0.84 \mathrm{~cm}^{-1}\)
- A \(6.0 \times 10^{-5} \Omega^{-1} \mathrm{~cm}^{-1}\)
- B \(3.0 \times 10^{-5} \Omega^{-1} \mathrm{~cm}^{-1}\)
- C \(9.0 \times 10^{-5} \Omega^{-1} \mathrm{~cm}^{-1}\)
- D \(12.0 \times 10^{-5} \Omega^{-1} \mathrm{~cm}^{-1}\)
Answer & Solution
Correct Answer
(A) \(6.0 \times 10^{-5} \Omega^{-1} \mathrm{~cm}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \mathrm{k} & =\frac{\text { cell constant }}{\mathrm{R}} \\ & =\frac{0.84 \mathrm{~cm}^{-1}}{14000 \Omega} \\ & =6.0 \times 10^{-5} \Omega^{-1} \mathrm{~cm}^{-1}\end{aligned}\)
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