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The specific conductivity of an aqueous solution of a weak monoprotic acid is 0.00033 ohm–1cm–1 at a concentration 0.02 M. If at this concentration the degree of dissociation is 0.043, then calculate the value of (in ohm–1 cm2 /eqt) :a) 483b) 438c) 348d) 384Correct answer is option 'D'. 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 specific conductivity of an aqueous solution of a weak monoprotic acid is 0.00033 ohm–1cm–1 at a concentration 0.02 M. If at this concentration the degree of dissociation is 0.043, then calculate the value of (in ohm–1 cm2 /eqt) :a) 483b) 438c) 348d) 384Correct answer is option 'D'. 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 specific conductivity of an aqueous solution of a weak monoprotic acid is 0.00033 ohm–1cm–1 at a concentration 0.02 M. If at this concentration the degree of dissociation is 0.043, then calculate the value of (in ohm–1 cm2 /eqt) :a) 483b) 438c) 348d) 384Correct answer is option 'D'. Can you explain this answer?.

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