When a certain conductance cell was finlled with 0.1 mol `L^(-1)` KCI solution, it had a resistance of 85 ohm at 298K. When the same cell was filled with an aqueous solution of 0.052 mol `L^(-1)` of an electrolyte, the resistance was 96 ohm. Calcualte the molar conductance of the electrolyte at this concentration. (Specific conductance of 0.1 mol `L^(-1)` KCI solution is 1.29xx10^(-2)ohm^(-1)cm^(-1))`.

When a certain conductance cell was finlled with 0.1 mol `L^(-1)` KCI

1.	When a certain conductance cell was finlled with 0.1 mol `L^(-1)` KCI solution, it had a resistance of 85 ohm at 298K. When the same cell was filled with an aqueous solution of 0.052 mol `L^(-1)` of an electrolyte, the resistance was 96 ohm. Calcualte the molar conductance of the electrolyte at this concentration. (Specific conductance of 0.1 mol `L^(-1)` KCI solution is 1.29xx10^(-2)ohm^(-1)cm^(-1))`.
Answer» Correct Answer - `220.2 S cm^(2)mol^(-1)` Step I. Calcualtion of cell constant. Resistance of KCl solution (R )=85 ohm Specific conductance, (k)`=1.29xx10^(-2) ohm^(-1) cm^(-1)` Specific conductance, (k)`=(1)/(R )xx"cell constant"` `:. " " "Cell constant" =kxxR=(1.29xx10^(-2) ohm^(-1)cm^(-1))xx85" ohm" =1.1 cm^(-1)` Step II. Calculation of molar conductance. Resistance of given electrolyte solution, (R )=96 ohm Specific conductance,(k)`=("Cell constant")/(R )=((1.1 "cm"^(-1)))/((96" ohm"))=1.145xx10^(-2)ohm^(-1)cm^(-1)` Concentration, C`=0.052 mol L^(-1)=(0.052" mol")/(1L)=(0.052" mol")/(1000" cm"^(3))=5.2xx10^(-5)" mol cm"^(3)` Molar conductance, `(Lambda_(m))=(k)/(C )=((1.145xx10^(-2) ohm^(-1) cm^(-1)))/((5.2xx10^(-5) mol cm^(-3)))` `=220.2 ohm^(-1) cm^(-1)=220.2 cm^(2) mol^(-1)`.

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