Course Content
CHAPTER 10: HALOALKANES AND HALOARENES
Topic Name 10 Haloalkanes and Haloarenes 10.1 Classification 10.2 Nomenclature 10.3 Nature of C–X Bond 10.4 Methods of Preparation of Haloalkanes 10.5 Preparation of Haloarenes 10.6 Physical Properties 10.7 Chemical Reactions
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CHAPTER 11: ALCOHOLS, PHENOLS AND ETHERS
Topic Name 11 Alcohols, Phenols and Ethers 11.1 Classification 11.2 Nomenclature 11.3 Structures of Functional Groups 11.4 Alcohols and Phenols 11.5 Some Commercially Important Alcohols 11.6 Ethers
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Class 12th Chemistry Online Class For 100% Result
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Arrangement for measurement of Resistance

Class 12 Electrochemistry Notes
  • The arrangement consists of two resistances R3 and R4.
  • There is a variable resistance R1 and a conductivity cell with unknown resistance R2.
  • The Wheatstone bridge is provided with an oscillator O that acts as source of a.c. power.
  • The arrangement has a suitable detector P.
  • The Wheatstone bridge is balanced when there is no flow of current through the detector.

Unknown resistance = R2 = R1R4 /R3

  • After calculating the resistance the conductiviry can be easily calculated using the formula:
  • κ = G’ /R

Molar conductivity

  • It is denoted by the symbolIt is related to the conductivity of the solution by the following equation:
Class 12 Electrochemistry Notes
  • The units of is S mmol-1.

Problem:

THE CONDUCTIVITY OF 0.20 M SOLUTION OF KCL AT 298 K IS 0.0248 SCM-1. CALCULATE ITS MOLAR CONDUCTIVITY.

Solution:

k = 0.0248 S cm – 1

c = 0.20 M

Molar conductivity,Λm =  (k x 1000) / c

= 0.0248 x1000 / 0.20

= 124 Scmmol – 1

Variation of Conductivity and Molar Conductivity with Concentration

  • They depend on the concentration of the electrolyte. The Conductivity and Molar Conductivity of both weak and strong electrolytes decreases withdecrease in concentration as the number of ions per unitvolume carrying the current in a solution decreases on dilution.
  • Conductivity of a solution at a specific concentration = Conductance of solution placed in between the two platinum electrodes where
  • Volume of solution = 1 unit

 Cross sectional area of electrodes = 1unit

Class 12 Electrochemistry Notes
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