MHT CET · Chemistry · Electrochemistry
The limiting molar conductivities \(\left(\Lambda_0\right)\) for NaCl , KBr and KCl are 126, 152 and \(150 \mathrm{~S} \mathrm{~cm} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\) respectively. What is the \(A_0\) of NaBr ?
- A \(128 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- B \(302 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- C \(278 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- D \(176 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(128 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \wedge_{0(\mathrm{NaBr})} & =\wedge_{0(\mathrm{NaCl})}+\wedge_{0(\mathrm{KBr})}-\wedge_{0(\mathrm{KCl})} \\ & =126+152-150 \\ & =128 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{aligned}\)
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