Exams › JEE Main › Chemistry
A 0.01 N solution of an electrolyte placed in a conductivity cell of cell constant 0.88 cm⁻¹ shows a resistance of 210 ohm at 298 K. What are the specific conductance (kappa) and the equivalent conductance of the solution?
- kappa = 4.19 * 10⁻³ S/cm; equivalent conductance = 419 S cm² eq⁻¹
- kappa = 4.19 * 10⁻³ S/cm; equivalent conductance = 41.9 S cm² eq⁻¹
- kappa = 1.86 * 10⁻³ S/cm; equivalent conductance = 186 S cm² eq⁻¹
- kappa = 0.88 * 10⁻³ S/cm; equivalent conductance = 88 S cm² eq⁻¹
Correct answer: kappa = 4.19 * 10⁻³ S/cm; equivalent conductance = 419 S cm² eq⁻¹
Solution
Conductance G = 1/R, and kappa = G * (cell constant) = (cell constant)/R = 0.88/210 = 4.19*10⁻³ S/cm. Equivalent conductance converts kappa to per equivalent: lambda_eq = kappa*1000/N = 4.19*10⁻³*1000/0.01 = 419 S cm² eq⁻¹.
Related JEE Main Chemistry questions
⚔️ Practice JEE Main Chemistry free + battle 1v1 →