Home
Class 12
CHEMISTRY
The resistance of 0.5 N solution of an e...

The resistance of 0.5 N solution of an electrolyte in a conductivity cell was found to be 25 ohm. Calculate the equivalent conductivity of the solution if the electrodes in the cell are 1.6 ci apart and have an area of 3.2 `cm^2` ?

A

` 10 S cm^2 eq^(-1)`

B

`15 cm^(2) eq^(-1)`

C

`20 S cm^(2) eq^(-1)`

D

`40 S cm^(2) e`

Text Solution

Verified by Experts

The correct Answer is:
D

`40 S cm^(2) e`
Promotional Banner

Topper's Solved these Questions

  • PHYSICAL CHEMISTRY OF SOLUTIONS

    MTG-WBJEE|Exercise WB JEE WORKOUT (Single Option Correct Type)(2 mark)|15 Videos
  • PHYSICAL CHEMISTRY OF SOLUTIONS

    MTG-WBJEE|Exercise WB JEE WORKOUT (One or More than One Option Correct Type )(2 mark)|10 Videos
  • MODEL TEST PAPTER

    MTG-WBJEE|Exercise MCQs|80 Videos
  • POLYMERS

    MTG-WBJEE|Exercise WB JEE PREVIOUS YEARS QUESTIONS (CATEGORY 1 : Single Option Correct Type )|2 Videos

Similar Questions

Explore conceptually related problems

(a). The resistance of a decinormal solution of an electrolyte in a conductivity cell was found to be 245 ohms. Calculate the equilvalet conductivity of the solution if the electrodes in the cell were 2 cm apart and each has an area of 3.5 sq. cm. (b). The conductivity of 0.001028M acetic acid is 4.95xx10^(-5)Scm^(-1) . Calculate its dissociation constant if ^^_(m)^(@) for acetic acid is 390.5Scm^(2_mol^(-1)

The resistance of a decinormal solution of an electrolyte in a conductivity cell was found to be 245 Omega . Calculate the equivalent conductance of the solution if the electrodes in the cell were 2 cm part and each had an area of 3.5 sq. cm.

The resistance of a 0.5 M solution of an electrolyte was found to be 30 Omega enclosed between two platinum electrodes. Calculate the molar conductivity of the solution if the electrodes in the cell are 1.5 cm apart and having an area of cross section 2.0 cm^(2) ?

The resistance of aN//10KCI solution is 245 Omega . Calculate the equivalent conductance of the solution if the electrodes in the cell are 4 cm apart and each haveing an area of 7.0 sq.cm .

The resistance of an N//10 KCl solution is 245 ohms . Calculate the equivalent conducatnce of the solution id the electrodes in the cell are 4 cm apart and each having an area of 7.0 sq.cm.

The resistance of 0.5 N solution of an electroolyte in a conductivity cell was found to be 45 ohms. If the electrodes in the cell are 2.2 cm apart and have an area of 3.8cm^(2) then the equivalent conductance (in Scm^(2)eq^(-1) ) of a solution is

The resistance of 0.01 N solution at 25^(@)C is 200 ohm. Cell constant of the conductivity cell is unity. Calculate the equivalent conductance of the solution.

The resistance of a N//10 KCI solution is 245 ohms. Calculate the specific conductance and the equivalent conductance of the solution if the electrodes in the cell are 4 cm apart and each having an area of 7.0 sq cm.