The resistance of a conductivity cell containing 0.01 {\M} {\KCl} solution at 298 {\K} is 1750 Ω. If the conductivity of 0.01 {\M} {\KCl} solution at 298 {\K} is 0.152 \times 10^{-3} {\S} {\cm}^{-1}, then the cell constant of the conductivity cell is _____\times 10^{-3} {\cm}^{-1} [24-Jun-2022-Shift-2]

JEE Main 24 June 2022 Shift 2 Solved Paper

Section: Chemistry

Question : 84 of 90

Marks: +1, -0

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

[24-Jun-2022-Shift-2]

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