CUET · CHEMISTRY · PYQ PAPER 2023
Limiting molar conductivity of \(H_2O\) is equal to :
- A \(\Lambda_m^{\circ} NaCl +\Lambda_m^{\circ} HCl -\Lambda_m^{\circ} NaOH\)
- B \(\Lambda_m^{\circ} HCl +\Lambda_m^{\circ} NaOH -\Lambda_m^{\circ} NaCl\)
- C \(\Lambda_m^{\circ} HCl +\Lambda_m^{\circ} NH _4 OH -\Lambda_m^{\circ} NaCl\)
- D \(\Lambda_m^{\circ} NaCl +\Lambda_m^{\circ} HCl -\Lambda_m^{\circ} NH _4 OH\)
Answer & Solution
Correct Answer
(B) \(\Lambda_m^{\circ} HCl +\Lambda_m^{\circ} NaOH -\Lambda_m^{\circ} NaCl\)
Step-by-step Solution
Detailed explanation
\(\Lambda_m^{\circ} HCl + \Lambda_m^{\circ} NaOH - \Lambda_m^{\circ} NaCl\) \(= (\lambda^{\circ} H^+ + \lambda^{\circ} Cl^-) + (\lambda^{\circ} Na^+ + \lambda^{\circ} OH^-) - (\lambda^{\circ} Na^+ + \lambda^{\circ} Cl^-)\) \(= \lambda^{\circ} H^+ + \lambda^{\circ} OH^-\)…
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