Exams › JEE Main › Chemistry
A 1 N salt solution has a resistance of 50 ohm when measured in a conductivity cell whose two platinum electrodes, each of area 4.2 cm², are 2.1 cm apart. Determine the equivalent conductance of this solution.
- 200 ohm⁻¹ cm² eq⁻¹
- 10 ohm⁻¹ cm² eq⁻¹
- 100 ohm⁻¹ cm² eq⁻¹
- 20 ohm⁻¹ cm² eq⁻¹
Correct answer: 10 ohm⁻¹ cm² eq⁻¹
Solution
The cell constant converts the measured conductance into specific conductance (conductivity). Dividing by normality and scaling by 1000 (to convert per litre to per cm³ basis) yields the equivalent conductance.
Related JEE Main Chemistry questions
⚔️ Practice JEE Main Chemistry free + battle 1v1 →