A cell with `N//50` KCl solution offered a resistance of 550 ohm at 298 K. The specific conductance of `N//50` KCl at 298 K is 0.002768 `ohm^(-1)cm^(-1)`. When this cell is filled with `N//10 ZnSO_(4)` solution, it offered a resistance of 72.18 ohm at 298 K. Find the cell constant and molar conductance of `ZnSO_(4)` solution at 298 K.

Answer

`1.522 cm^(-1), 4.22xx10^(2) S cm^(2)mol^(-1)`

Explanation

Step I. Calculation of cell constant
`R=550" ohm",k=0.002768" ohm"^(-1)cm^(-1)=2.768xx10^(-3) ohm^(-1)cm^(-1)`
Cell constant`(Lambda_(m)) =(kxx1000)/(M)`
Specific conductance(k)`=("cell constant")/(R )=((1.522 cm^(-1)))/((72.18" ohm"))=2.11xx10^(-2)ohm^(-2)ohm^(-1)cm^(-1)`
Molarity of solution (M)`=(0.1)/(2)=0.05 M " "(`:'`Valency of Zn^(2+)=2)`
Molar conductance`(Lambda_(m))=((2.11xx10^(-2)ohm^(-1)cm^(-1))xx(1000cm^(3)))/((0.05" mol"))`
`=422" ohm"^(-1)cm^(2)mol^(-1)=4.22xx10^(2)" S ""cm"^(2)" mol"^(-1)`.