Q. The specific conductivity of 0.1 N KCl solution is 0.0129 $\mathrm{\Omega}^{- 1}{cm}^{- 1}.$ The resistance of the solution in the cell 100Ω. The cell constant of the cell will be
A) 1.10
B) 1.29
C) 0.56
D) 2.80
Explanation
(b)
Specific conductivity $(\kappa)$
$= \ \frac{1}{R}\ \times cell\ constant$
$Cell\ constant = \kappa\ \times R$
= 0.0129 × 100 = 1.29
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.