For KCl solution, R = 1750 ohms
Conductivity, K =0.152×10−3Scm−1
K=(al)R1
Cell constant,al=KXR=0.152×10−3X1750=266X10−3cm−1
JEE Main 2022 — Chemistry Physical Chemistry
The resistance of a conductivity cell containing 0.01MKCl solution at 298K is 1750Ω. If the conductivity of 0.01MKCl solution at 298K
is 0.152×10−3Scm−1, then the cell constant of the conductivity cell is____×10−3cm−1
Held on 24 Jun 2022 · Verified 6 Jul 2026.
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