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JEE Main Chemistry question

ChemistryElectrochemistryJEE Main

The limiting molar conductivities for NaCl, KBr and KCl are 126, 152 and 150 respectively. The for NaBr is

Answer: (A)

Solution

By Kohlrausch's law, Missing superscript or subscript argumentΛ^\circ_{NaBr}Missing superscript or subscript argument=Λ^\circ_{NaCl}Missing superscript or subscript argument+Λ^\circ_{KBr}Missing superscript or subscript argument-Λ^\circ_{KCl}$=126+152

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Question ID VK-143693