MHT CET · Chemistry · Electrochemistry
The conductivity of 0.02 M KCl , solution is \(0.00250 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\). What is its molar conductivity?
- A \(125 \mathrm{ohm}^{-1} \mathrm{~cm}^{-2} \mathrm{~mol}^{-1}\)
- B \(0.05 \mathrm{ohm}^{-1} \mathrm{~cm}^{-2} \mathrm{~mol}^{-1}\)
- C \(725 \mathrm{ohm}^{-1} \mathrm{~cm}^{-2} \mathrm{~mol}^{-1}\)
- D \(8000 \mathrm{ohm}^{-1} \mathrm{~cm}^{-2} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(125 \mathrm{ohm}^{-1} \mathrm{~cm}^{-2} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\text {Molar conductivity}=\frac{1000 \mathrm{k}}{\mathrm{c}} \)
\( =\frac{1000 \mathrm{~cm}^3 \mathrm{~L}^{-1} \times 0.00250 \Omega^{-1} \mathrm{~cm}^{-1}}{0.02 \mathrm{molL}^{-1}} \)
\( = 125 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
\( =\frac{1000 \mathrm{~cm}^3 \mathrm{~L}^{-1} \times 0.00250 \Omega^{-1} \mathrm{~cm}^{-1}}{0.02 \mathrm{molL}^{-1}} \)
\( = 125 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
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