A cell is constructed having Pt electrodes of area `5cm^(2)` and separated by a distance of 20cm .The resistance of a 0.1M solution is found to be` 50Omega` ,the specific conductance of the solution in Sl units is

The resistance of 0.0025M solution of K_(2)SO_(4) is 326ohm. The specific conductance of the solution, if cell constant is 4.

The resistance of decinormal solution is found to be 2.5 xx 10^(3) Omega . The equivalent conductance of the solution is (cell constant = 1.25 cm^(–1) )

The resistance of 0.1 N solution of a salt is found to be 2.5xx10^(3) Omega . The equivalent conductance of the solution is (Cell constant =1.15 cm^(-1) )

The resistance of 0.01N solution of an electrolyte AB at 328K is 100ohm. The specific conductance of solution is (cell constant = 1cm^(-1) )

A 0.1 M solution of monobasic acid at specific resistance of r ohms-cm, its molar conductivity is

At 298K the resistance of a 0.5N NaOH solution is 35.0 ohm. The cell constant is 0.503 cm^(-1) the electrical conductivity of the solution is

If resistivity of 0.8M KCl solution is 2.5xx10^(3)Omega cm calculate lamda_(m) of the solution

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.

A solution of 0.1 N KCl offers a resistance of 245 ohms. Calculate the specific conductance and the equivalent conductance of the solution if the cell constant is 0.571 cm^(-1) .

Resistance of 0.2 M solution of an electrolyte is 50Omega . The specific conductance of the solution is 1.4 Sm^(-1) . The resistance of 0.5 M solution of the same electrolyte is :