Resistance of a conductivity cell (cell constant 129 m - 1 ) filled with 74 . 5 ppm solution of KCl is 100 Ω (labelled as solution 1 ). When the same cell is filled with KCl solution of 149 ppm , the resistance is 50 Ω (labelled as solution 2 ). The ratio of molar conductivity of solution 1 and solution 2 is i.e. ∧ 1 ∧ 2 = x × 10 - 3 . The value of x is____Given, molar mass of KCl is 74 . 5

Step-by-step solution

Given l A = 129   m - 1 KCl solution 1 : 74 . 5 ppm ,       R 1 = 100   Ω KCl solution 2 : 149 ppm ,         R 2 = 50   Ω Here, ppm 1 ppm 2 = M 1 M 2 and κ   =   1 R G *       where   G *   =   cell   constant Since value of cell constant is same in both the cases. k 1 k 2   =   R 2 R 1 ∧ 1 ∧ 2 = k 1 × 1000 M 1 k 2 × 1000 M 2 = K 1   K 2 × M 2 M 1   =   R