Home
Class 12
CHEMISTRY
The specific conductivity of 0.12N solut...

The specific conductivity of 0.12N solution of an electrolyte is `2.4 xx 10^(-2)S cm^(-1)`. Calculate its equivalent conductance.

Text Solution

Verified by Experts

`200s cm^(2) eq^(-1)`
Doubtnut Promotions Banner Mobile Dark
|

Similar Questions

Explore conceptually related problems

The specific conductance of a 0.12N solution of an electrolyte is 2.4xx10^-2 ohm^-1cm^-1 Calculate its equivalent conductance.

Specific conductance of a 0.12 N solution of electrolyte is 2.4 xx 10^(-2) "ohm^(-1) cm^(-1) . What is the value of its equivalent conductance?

The resistance of a conductivity cell when filled with 0.05M solution of an electrolyte x is 100 Omega at 40^(@)C . The same conductivity cell filled with 0.01M solution of electrolyte y has a resistance of 50 Omega . The conductivity of 0.05 M solution of electrolyte x is 1.0 xx 10^(-4) S cm^(-1) . Calculate (i) cell constant (ii) conductivity of 0.01 M solution (iii) molar conductivity of 0.01 M solution

The conductivity of a 0.20 M solution of KCl at 298 K is 0.0248 S cm^(-1) . Calculate its molar conductivity.

The conductivity of 0.20 M solution of KCl at 298 K is 0.248S cm^(-1) . Calculate its molar conductivity.

The conductivity of a 0.20 M solution of KCl at 298K is 0.0248S cm^(-1) . Calculate its molar conductivity.

A 0.1N solution of NaCl has specific conductance 0.0011 ohm^(-1) cm^(-1) . Find its equivalent conductance.

At a certain temperature the specific conductance of 0.40M KCl solution is 4.96xx10^-2 ohm^-1cm^-1 . Calculate its molar conductance.

molar conductance of 1.5 M solution of an electrolyte is 138.9 S cm^(2) Mol^(-1) . Find the specific conductance.

The resistance of N//5 solution of an electrolyte in a cell was found to be 45 ohm. Calculate the equivalent conductance of the solution if the electrodes in a cell are 2.2 cm apart and have an area of 3.8 equiv. cm.

Home
Profile