MHT CET · Chemistry · Electrochemistry
If resistivity of \(0.8 \mathrm{M} \mathrm{KCl}\) solution is \(2 \cdot 5 \times 10^{-3} \Omega \mathrm{cm} .\) Calculate molar conductivity of solution?
- A \(3 \times 10^{5} \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
- B \(2 \times 10^{5} \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
- C \(4 \times 10^{5} \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
- D \(5 \times 10^{5} \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(D) \(5 \times 10^{5} \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
(A)
\(\text {Conductivity}(\mathrm{k})=\frac{1}{\text {resistivity}}=\frac{1}{2.5 \times 10^{-3} \Omega \mathrm{cm}}=\) \(0.4 \times 10^{3} \Omega^{-1} \mathrm{~cm}^{-1} \)
\( \text {Molar conductivity} \quad\left(\Lambda_{\mathrm{m}}\right)=\frac{1000 \mathrm{k}}{\mathrm{C}}=\) \(\frac{1000 \mathrm{~cm}^{3} \mathrm{~L}^{-1} \times 0.4 \times 10^{3} \Omega^{-1} \mathrm{~cm}^{-1}}{0.8 \mathrm{~mol} \mathrm{~L}^{-1}}\)
\(\text {Conductivity}(\mathrm{k})=\frac{1}{\text {resistivity}}=\frac{1}{2.5 \times 10^{-3} \Omega \mathrm{cm}}=\) \(0.4 \times 10^{3} \Omega^{-1} \mathrm{~cm}^{-1} \)
\( \text {Molar conductivity} \quad\left(\Lambda_{\mathrm{m}}\right)=\frac{1000 \mathrm{k}}{\mathrm{C}}=\) \(\frac{1000 \mathrm{~cm}^{3} \mathrm{~L}^{-1} \times 0.4 \times 10^{3} \Omega^{-1} \mathrm{~cm}^{-1}}{0.8 \mathrm{~mol} \mathrm{~L}^{-1}}\)
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