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MHT CET · Chemistry · Electrochemistry

What will be the concentration of NaCl solution, if the molar conductivity and conductivity of NaCl solution is \(124 \cdot 3 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) and \(1 \cdot 243 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{2}\)
respectively?

  1. A \(0.001\) mol \(\mathrm{L}^{-1}\)
  2. B \(0.01 \mathrm{~mol} \mathrm{~L}^{-1}\)
  3. C \(0.02\) mol \(I_{1}^{-1}\)
  4. D \(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(0.001\) mol \(\mathrm{L}^{-1}\)

Step-by-step Solution

Detailed explanation

(D)
\(\Lambda=124.3 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}, \mathrm{k}=1.243 ~\times\) \(10^{-4} \Omega^{-1} \mathrm{~cm}^{2} \)
\( \mathrm{C}=? \)
\( \wedge=\frac{1000 \mathrm{k}}{\mathrm{C}} \quad \therefore \mathrm{C}=\frac{1000 \mathrm{k}}{\wedge} \)
\( \therefore \mathrm{C}=\frac{1000 \times 1.243 \times 10^{-4}}{124.3}=0.001 \mathrm{~mol} \mathrm{~L}^{-1}\)