Question

The molar conductivity of a 0.5mol/dm3 solution of AgNO3 with electrolytic conductivity of 5.76×10−3S cm−1 at 298 K is

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

Solution

The correct option is D 11.52 S cm2/molC=0.5 mol/dm3 k=5.76×10−3 Scm−1 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 =5.76×10−3×10+30.5=11.52 S cm2/mol

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