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

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

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

Specific conductance of 0.1M nitric acid is 6.3xx10^(-2)ohm^(-1)cm^(-1) . The molar conductance of the solution is:

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.

Resistance of a 0.1 M KCl solution in a conductance cell is 300 ohm and specific conductance of "0.1 M KCl" is 1.33xx10^(-2)" ohm"^(-1)"cm"^(-1) . The resistance of 0.1 M NaCl solution in the same cell is 400 ohm. The equivalent conductance of the 0.1 M NaCl solution ("in ohm"^(-1)"cm"^(2)"/gmeq.") is

In a conductivity cell the two platinum electrodes each of area 10 sq. cm are fixed 1.5 cm apart. The cell. Contained 0.05N solution of a salt. If the two electrodes are just half dipped into the solution which has a resistance of 50 ohms, find equivalent conductance of the salt solution.