JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
The resistivity of a \(0.8\,M\) solution of an electrolyte is \(5 \times 10^{-3}\,\Omega\,cm\). Its molar conductivity is \(.....\times 10^4 \Omega^{-1}\,cm ^2\,mol ^{-1}\). (Nearest integer)
- A \(24\)
- B \(23\)
- C \(25\)
- D \(22\)
Answer & Solution
Correct Answer
(C) \(25\)
Step-by-step Solution
Detailed explanation
\(\Lambda_{ m }= \frac{\kappa \times 1000}{ M }\) \(\Lambda_{ m }= \frac{1}{\rho} \times \frac{1000}{ M }\) \(\frac{1}{5 \times 10^{-3}} \times \frac{1000}{0.8}\) Ans. \(25 \times 10^4\,\Omega^{-1}\, cm ^{-2}\, mol ^{-1}\)
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