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NEET · Chemistry · STD 12 - 2. Electrochemistry
Given that \(\Lambda_{ m }^{\alpha}=133.4\,\left( AgNO _{3}\right) ; \Lambda_{ m }^{\alpha}=149.9( KCl )\) \(\Lambda_{ m }^{\alpha}=144.9\, S\, cm ^{2} \,mol ^{-1}\left( KNO _{3}\right)\) the molar conductivity at infinite dilution for \(AgCl\) is \(.......\, S \,cm ^{2}\, mol ^{-1}\)
- A \(140\)
- B \(138\)
- C \(134\)
- D \(132\)
Answer & Solution
Correct Answer
(B) \(138\)
Step-by-step Solution
Detailed explanation
\(\wedge_{ m }^{0}( AgCl )=\wedge_{ m }^{0}\left( AgNO _{3}\right)+\wedge_{ m }^{0}( KCl )-\wedge_{ m }^{0}\left( KNO _{3}\right)\) \(=133.4+149.9-144.9\)
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