The molar conductivity of 0.04 M solution of `MgCl_2` is 200 `Scm^3mol^(-1)` at 298 k. A cell with electrodes that are 2.0`cm^2` in surface area and 0.50cm apart is filled with `MgCl_2` solution
How much current will flow when the potential difference between the two electrodes is 5.0V?

The molar conductivity of 0.04 M solution of MgCl_2 is 200 Scm^2mol^(-1) at 298 k. A cell with electrodes that are 2.0 cm^2 in surface area and 0.50cm apart is filled with MgCl_2 solution. Conductance of MgCl_2 solution is :

The conductivity of 0.20 MKCl solution at 298 K is 0.0248 S cm^-1 . Calculate its molar conductivity.

The conductivity of 0.35M NaCl solution at 298K is 0.025 S/cm. Calculate its molar conductivity .

The molar conductivity of 1.5 M solution of an electrolysis is found to be 138.9 S cm^2 mol^-1 . Calculate the conductivity of this solution?

The molar conductivity of a 0.5 mol/dm3 solution of AgNO3 with electrolytic conductivity of 5.76 x 10-3 S cm-1 at 298 K is

The conductivity of a solution containing 1.0 g of anhydrous BaCl_2 in 200 cm^3 of the solution has been found to be 0.0058 S cm^-1 . Calculate the molar conductivity and equivalnet conductivity of the solution.

The resistance of a conductivity cell containing 0.001M KCl solution at 298K is 1500 Omega . What is the cell constant ( in mm^(-1) ) if the conductivity of 0.001M KCl solution is 2xx10^(-3)Smm^(-1)

The resistance of a conductivity cell with 0.1 M KCl solution is found to be 200 ohm at 298K. When the same cell was filled with 0.02 M NaCl solution, the resistance at the same temperature is found to be 1100 ohm . Calculate: the cell constant of the cell in m^-1 . Given conductivity of 0.lM KCl is 1.29 Sm^(-1) ]

A battery 9 V is connected in series with resistors of 0.2 Omega ,0.3 Omega 0.4 Omega 0.5 Omega and 12 Omega respectively.How much current will flow through a 12 Omega resistor?