The distance between two electrodes of a cell is 2.5cm and area of each electrode is `5cm^(2)` the cell constant (in `cm^(-1)`)

The distance between the eletrodes of a conductivity cell is 0.98 cm and the cell constant is 0.5 cm^(-1) . Calculate the cross-sectional area of electrode.

The conductance of 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross-section of 1 cm^(2) . The conductance of this solution was found to be 5xx10^(-7) S . The pH of the solution is 4. Calculate the value of limiting molar conductivity.

Twoelectrodes are fitted in conductance cell 1.5 cm apart while the area of cross-section of each electrode is 0.75 cm^(2) . The cell constant is