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The resistance of 0.01 N solution of an electrolyte was found to be 210 ohm at 298 K, using a conductivity cell of cell constant 0.66 cm−1. The equivalent conductance of the solution is
  • a)
    3.14 mho−2cm2eq−1
  • b)
    314.28 mho−1cm2eq−1
  • c)
    31.4 mho cm2eq−1
  • d)
    314.29 mho cm2eq−1
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The resistance of 0.01 N solution of an electrolyte was found to be 21...
Hence, the equivalent conductance of the solution is 314.29 Scm2eq−1 or 314.29 mho cm2eq−1.
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The resistance of 0.01 N solution of an electrolyte was found to be 21...
To find the resistivity (ρ) of the solution, we can use the formula:

ρ = (R * A) / L

Where:
ρ = resistivity of the solution
R = resistance of the solution (210 ohm)
A = cross-sectional area of the conductivity cell (0.66 cm^2)
L = distance between the electrodes of the conductivity cell (1 cm, since it is not specified)

Substituting the given values into the formula:

ρ = (210 ohm * 0.66 cm^2) / 1 cm
ρ = 138.6 ohm * cm^2 / cm
ρ = 138.6 ohm * cm

Therefore, the resistivity of the 0.01 N solution of the electrolyte at 298 K is 138.6 ohm * cm.
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The resistance of 0.01 N solution of an electrolyte was found to be 210 ohm at 298 K, using a conductivity cell of cell constant 0.66 cm−1. The equivalent conductance of the solution isa)3.14 mho−2cm2eq−1b)314.28 mho−1cm2eq−1c)31.4 mho cm2eq−1d)314.29 mho cm2eq−1Correct answer is option 'D'. Can you explain this answer?
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The resistance of 0.01 N solution of an electrolyte was found to be 210 ohm at 298 K, using a conductivity cell of cell constant 0.66 cm−1. The equivalent conductance of the solution isa)3.14 mho−2cm2eq−1b)314.28 mho−1cm2eq−1c)31.4 mho cm2eq−1d)314.29 mho cm2eq−1Correct 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 resistance of 0.01 N solution of an electrolyte was found to be 210 ohm at 298 K, using a conductivity cell of cell constant 0.66 cm−1. The equivalent conductance of the solution isa)3.14 mho−2cm2eq−1b)314.28 mho−1cm2eq−1c)31.4 mho cm2eq−1d)314.29 mho cm2eq−1Correct 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 resistance of 0.01 N solution of an electrolyte was found to be 210 ohm at 298 K, using a conductivity cell of cell constant 0.66 cm−1. The equivalent conductance of the solution isa)3.14 mho−2cm2eq−1b)314.28 mho−1cm2eq−1c)31.4 mho cm2eq−1d)314.29 mho cm2eq−1Correct answer is option 'D'. Can you explain this answer?.
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