First slide

Applications of Kohlraush law

Question

Molar conductance of an electrolyte A₃B₂ is ‘X’ mho cm² mole^-1. Equivalent conductance of A₃B₂ will be . . . mho cm² geq^-1

Easy

A

3X
B

6X
C

X/6
D

X/3

Solution

$\land_{eq} \times valency = \land_{m} \land_{eq} (A_{3} B_{2}) = \frac{X}{6} valency of A_{3} B_{2} is six A_{3} B_{2} ⇌ 3 A^{+ 2} + 2 B^{- 3} Net charge of anion = Net charge of cation = 6$

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