The molar conductivity of a 0.5mol/dm3 solution of AgNO3 with electrolytic conductivity of 5.76×10−3Scm−1 at 298 K is
A
0.086Scm2/mol
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B
28.8Scm2/mol
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C
2.88Scm2/mol
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D
11.52Scm2/mol
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Solution
The correct option is D11.52Scm2/mol C=0.5mol/dm3 k=5.76×10−3Scm−1 What this indicates is that if we have a cell of 1 mL electrolyte solution, its conductivity will be equal to k.
T=298K
λm is conductivity of 1 mole of the electrolyte. That is why to obtain it we divide k by concentration to obtain the value for 1 mole and multiply it with 1000 (to extrapolate to 1 L from 1 mL).