Q1.When a certain conductance cell was filled with 0.20 mol dm-3 aqueous KCl solution , its conductivity was 2.78 x 10-3 S.cm-1 and had a resistance 82.5 ohm at 300K.Calculate the cell  constant .
Q2. Calculate the resistance of 0.1N solution of an electrolyte whose equivalent conductivity is 420 Ω-1.cm2.equi-1. The cell constant is 0.88cm-1.
Q3. Molar conductance of 1.5M solution of an electrolyte is found to be 13.895 S.m2.mol-1. What would be the specific conductivity for this solution.
Q 4.The resistance of 0.5M solution of an electrolyte enclosed between platinum electrodes 1.5cm apart and having an area of 2.0cm2 was found to be 30 Ω .Calculate the molar conductivity of the solution.
Q 5.When a certain conductance cell was filled with 0.1molL-1 KCl, it has a resistance of 85 Ω at 250C. When the same cell was filled with an aqueous solution of 0.52 molL-1 of an electrolyte solution ,the resistance was 96 Ω .Calculate the molar conductivity of the electrolyte at this concentration. (Conductivity ? of 0.1molar KCl is 1.29x10-2S.cm)







Asked by Lochana Jayachandran | 20th Jul, 2014, 10:57: AM

Expert Answer:

Dear geetha0261@gmail.com

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.

We cannot entertain more than 5 questions in a subject per day. In case of multiple questions within a query, please post each question individually and let us know where you are getting stuck so that we would be able to explain things better.

Solution for your first query,

Cell constant = G* = Resistance x Conductivity

= 82.5 ohm x 2.78 x 10-3 S.cm-1

= 0.229 cm-1


Topperlearning Team.

Answered by Prachi Sawant | 21st Jul, 2014, 11:50: AM

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