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Question

The molar conductivity of a 0.5mol/dm3 solution of AgNO3 with electrolytic conductivity of 5.76×103S cm1 at 298 K is


Your Answer
A
0.086 S cm2/mol
Your Answer
B
28.8 S cm2/mol
Your Answer
C
2.88 S cm2/mol
Correct Answer
D
11.52 S cm2/mol

Solution

The correct option is D 11.52 S cm2/mol
C=0.5 mol/dm3
k=5.76×103 Scm1
What this indicates is that if we have a cell of 1 mL electrolyte solution, its conductivity will be equal to k

T=298 K

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).

λm=k×1000M
=5.76×103×10+30.5=11.52 S cm2/mol

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