MHT CET · Chemistry · Electrochemistry
Calculate molar conductivity at infinite dilution for \(\mathrm{NaBr}\) if molar conductivity at infinite dilution for \(\mathrm{NaCl}, \mathrm{KBr}\) and \(\mathrm{KCl}\) are 126, 152 and \(150 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\) respectively
- A \(128 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- B \(302 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- C \(278 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- D \(176 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(128 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \wedge{ }^0 \mathrm{NaBr}=\wedge{ }^0 \mathrm{NaCl}+\wedge^{\circ} \mathrm{KBr}-\wedge^{\circ} \mathrm{KCl} \\ & =126+152-150 \\ & =128 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mole}^{-1}\end{aligned}\)
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