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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 \(298 \mathrm{~K}\) ?

  1. A \(7.50 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
  2. B \(1 \cdot 333 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
  3. C \(1 \cdot 333 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}\)
  4. D \(1 \cdot 875 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(1 \cdot 333 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)

Step-by-step Solution

Detailed explanation

Conductivity, \(\mathrm{k}=\frac{\text { Cell constant }}{\mathrm{R}}=\frac{0.5 \mathrm{~cm}^{-1}}{375 \Omega}\)
\(\therefore \mathrm{k}=1.333 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)