The resistance of 0.1N solution of formic acid is 200 ohm and cell constant is 2.0 cm^(-1) . The equivalent conductivity ( in Scm^(2) eq^(-1)) of 0.1N formic acid is :
The resistance of 0.1N solution of acetic acid 250 ohm. When measured in a cell constant 1.15 cm^-1 . The equilvalent conductance (in ohm^-1 cm^2 eq^-1 ) of 0.1 N acetic acid is
Resistance of 0.2 M solution of an electrolyte is 50 Omega . The specific conductance of the solution is 1.3 Sm^-1 . If resistance of the 0.4 M solution of the same electrolyte is 260 Omega , its molar conductivity is:
The resistance of a 0.5 M solution of an electrolyte in a conductivity 'cell was found to be 25 ohm. Calculate the molar conductivity of the solution, if the electrodes in the cell are 1.6 cm apart and have an area of 3*2cm^(2) .
The conductivity of 0.1 n NaOH solution is 0.022 S cm^(-1) .When equal volume of 0.1 N HCl solution is added, the conductivity of resultant solution is decreased to 0.0055 S cm^(-1) . The equivalent conductivity of NaCl solution is :
The conductivity of 0.35M NaCl solution at 298K is 0.025 S/cm. Calculate its molar conductivity .
A 0.05 M NaOH solution offered a resistance of 31.6 ohm ina conductivity cell at 298 K. if the area of the plates of the conductivity cell is 3.8 cm^2 and distance between them 1.4 cm. calculate the molar conductivity of the NaOH solution.
The resistance of a 0.5M solution of an electrolyte was found to be 30Omega . Calculate the molar conductivity of the solution if the electrods in the cell are 1.5 cm a part and having an area of cross section is 2.0 cm^(2) .
The resistance of 0.05 M NaOH solution in a cell having length 5 cm area of cross section 10 cm2 is 5.55 xx 10^3 ohm. Calculate its molar conductance.
