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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 × 10-3 S cm-1, then the cell constant of the conductivity cell is _______ × 10-3 cm-1. (In integers)Correct answer is '266'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 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 × 10-3 S cm-1, then the cell constant of the conductivity cell is _______ × 10-3 cm-1. (In integers)Correct answer is '266'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 × 10-3 S cm-1, then the cell constant of the conductivity cell is _______ × 10-3 cm-1. (In integers)Correct answer is '266'. Can you explain this answer?.

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