VidhyaKosh
Test SeriesFree PYQsExamsPricing

JEE Main 2004 Chemistry question

ChemistryElectrochemistryJEE Main 2004

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

Answer: (1)

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}$ -

See the full step-by-step solution

Create a free account to read the complete working, and practise questions like this in a timed test.

Create a free account

Question ID VK-038408