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JEE Main 2022 · 24 Jun · Shift 2, Chemistry question 84

The resistance of a conductivity cell containing 0.01 M KCl solution at 298 K is 1750 Ω1750\ \Omega. If the conductivity of 0.01M KCl solution at 298 K is 0.152×10−3 S cm−10.152\times10^{-3}\ \mathrm{S\,cm^{-1}}, then the cell constant of the conductivity cell is ______ ×10−3 cm−1\times10^{-3}\ \mathrm{cm^{-1}}.

Answer and worked solution

Official answer: 266

Cell constant. G∗=κR=0.152×10−3×1750=0.266G^*=\kappa R=0.152\times10^{-3}\times1750=0.266 cm−1=266×10−3^{-1}=266\times10^{-3} cm−1^{-1}.

Answer: 266

Topics:Conductance and Kohlrausch's law
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