MHT CET · Chemistry · Electrochemistry
What is the conductivity of \(0.01\) M NaCl solution if resistance and cell constant of NaCl solution are 375 ohms and \(0.5 \mathrm{~cm}^{-1}\) respectively at \(298 \mathrm{~K}\) ?
- A \(7.50 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
- B \(1 \cdot 333 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
- C \(1 \cdot 333 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}\)
- D \(1 \cdot 875 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
Answer & Solution
Correct Answer
(B) \(1 \cdot 333 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
Step-by-step Solution
Detailed explanation
Conductivity, \(\mathrm{k}=\frac{\text { Cell constant }}{\mathrm{R}}=\frac{0.5 \mathrm{~cm}^{-1}}{375 \Omega}\)
\(\therefore \mathrm{k}=1.333 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
\(\therefore \mathrm{k}=1.333 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
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