JEE Main2004ChemistryElectrochemistryActual
The limiting molar conductivities ^ for NaCl , KBr and KCl are 126,152 and 150 ~S ~cm ^2 ~mol ⁻¹ respectively. The ^ for NaBr is
Options
- A128 ~S ~cm ^2 ~mol ⁻¹
- B302 ~S ~cm ^2 ~mol ⁻¹
- C278 ~S ~cm ^2 ~mol ⁻¹
- D176 ~S ~cm ^2 ~mol ⁻¹
Correct answer
A. 128 ~S ~cm ^2 ~mol ⁻¹
Step-by-step solution
_ NaCl ^ = _ Na ^ + _ Cl ^ =126 (1) _ KBr ^ = _ K ⁺ ^ + _ Br ⁻ ^ =152 _ KCl ^ = _ K ⁺ ^ + _ Cl ⁻ ^ =150 _ NaBr ^ = _ Na ^ + _ Br ⁻ ^ _ NaBr ^ =126+152-150=128