The conductance of a salt solution (AB) measured by two parallel electodes of area `100 cm^2` separated by 10cm was found to be `0.0001Omega^(-1)` . If volume enclosed between two electrode contain 0.1 mole of salt, what is the molar conductivity `(Scm^2mol^(-1))` of salt at same concentration:

The conductance of a salt solution (AB) measured by two parallel electrodes of area 100 cm^(2) separated by 10 cm was found to be 0.0001 Omega^(-1) . If volume enclosed between two electrode contains 0.1 mole of salt, and the molar conductivity (S cm^(2) mol^(-1)) of salt at same concentration is 1.0 xx 10^(-x) , x is

The resistance of a decinormal solution of a salt occupying a volume between two platinum electrodes which are 1.80 cm apart and 5.4 cm^(2) in area was found to be 50 omega calculate the equilvalent conductance of the solution

The conductivity of 0.01 mol L^(-1) KCl solution is 1.41xx10^(-3)" S "cm^(-1) . What is the molar conductivity ( S cm^(2) mol^(-1) ) ?

1N salt solution surrounding two platinum electrodes, 2.1 cm apart and 6.3 cm^(2) in area was found to offer a resistance of 50 ohm. Calculate the equivalent conductance of the solution.

One normal salt solution surrounding two platinum electrodes, 2.1 cm apart and 6.3 cm^(2) in area was found to offer a resistance of 50 ohm. Calculate the equivalent conductivity solution.

One normal salt solution surrounding two platinum electrodes, 2.1 cm apart and 6.3 cm^(2) in area was found to offer a resistance of 50 ohm. Calculate the equivalent conductivity solution.