The specific conductivity of a solution containing `1.0g` of anhydrous `BaCI_(2)` in `200 cm^(3)` of the solution has been found to be `0.0058 S cm^(-1)`. Calculate the molar and equivalent conductivity of the solution. Molecular wt. of `BaCI_(2) = 208`[mu implies `lambda_(m)]`

Answer

`241.67 S cm^(2)mol^(-1) ; 120.83 S cm^(2)(g eq.)^(-1)`

Explanation

Step I. Calculation of molarity (M) and normality (N) of the solution
Molarity (M)`=("Mass of" BaCl_(2)//"Molar mass")/("Volume of solution in litres")=((1g)//(208"g mol"^(-1)))/((200//1000" cm"^(3)))=0.024" mol cm"^(-3)`
Normality (N)`=0.024xx2=0.048" g eq" cm^(-3) " " (because "Valency of" Ba^(2+)=2)`
Step II. Calculation of molar conductivity of solution
`Lambda_(m)=(kxx1000)/(M)=((0.0058" ohm"^(-1)cm^(-1))xx(1000 cm^(3)))/((0.24" mol"))=241.67" ohm"^(-1) cm^(2)mol^(-1)`.
Step III. Calcualtion of molar conductivity of solution
`Lambda_(E)=(kxx1000)/(N)=((0.0058" ohm"^(-1)cm^(-1))xx(1000 cm^(3)))/((0.048" eq"))=120.83" ohm"^(-1) cm^(2)eq^(-1)`.