Chemistry · 2022
JEE Main · 24 June 2022, Shift 2 · Q84
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…
The resistance of a conductivity cell containing $\displaystyle 0.01$ M KCl solution at $\displaystyle 298$ K is $\displaystyle 1750 \Omega$. If the conductivity of $\displaystyle 0.01$ M KCl solution at $\displaystyle 298$ K is $\displaystyle 0.152 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1}$, then the cell constant of the conductivity cell is
\_\_\_\_ $\displaystyle \times 10^{-3} \mathrm{~cm}^{-1}$.
Official answer
266
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JEE Main 2022 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.