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Question

What will be the molar conductance ’Λ’, if resistivity is x for 0.1 N H2SO4?


A

Λ=x×1000/0.1

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B

Λ=2×1000/x×0.1

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C

Λ=x×1000/0.5

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D

Λ=0.5/1000x

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Solution

The correct option is B

Λ=2×1000/x×0.1


Explanation for correct option:

(B) Λ=2×1000/x×0.1

Step 1: Molar conductivity:

  • “The conductivity of solutions of different electrolytes in the same solvent and at a given temperature differs due to charge and size of the ions in which they dissociate, the concentration of ions, or ease with which the ions move under a potential gradient.”
  • "It, therefore, becomes necessary to define a physically more meaningful quantity called molar conductivity denoted by the symbol Λm."
  • It is related to the conductivity of the solution by the equation:

Λm=Kc, if K is expressed in Sm-1 and the concentration, c in molm-3 then the units of Λm are in Sm2mol-1.

  • Or it can also be expressed as, ΛmScm2mol-1=KScm-11000Lm-3×MolaritymolL-1

Step 2: Given information:

  • K = Specific conductance = 1 / specific resistance = 1 / 1
  • M = Molarity = 0.1 / 2

Step 3: Calculation for Molar conductance:

Λm=1000×(1/x)(0.1/2)Λm=2×1000x×0.1

Hence, the molar conductance will be, Λ=2×1000/x×0.1.

Explanation for incorrect options:

The molar conductance is Λ=2×1000/x×0.1. Thus, options A, C, and D are incorrect.

Hence, the correct option is B i.e. Λ=2×1000/x×0.1.


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