Resistance of ` 0.2 M` solution of an electrolue is ` 50 Omega`. The specific conductance of the solution is ` 1.4 S m^^(-1)`. The resistance of `0.5` M solution of the same electrolyte is `280. Omega`. The molar conducitivity of ` 0.5 M` solution of the electrolyte is ` S m^2 "mol"^(-1)` is.

Options

A. `5xx10^(2)`
B. `5xx10^(-4)`
C. `5xx10^(-3)`
D. `5xx10^(3)`

Answer

B

Explanation

Case I: `C=0.2M,R=50Omega,kappa=1.4" S "cm^(-1)`
`kappa`=Conductance`xx`Cell const.`=(1)/(R)xx`cell const. ltBrgt `therefore`Cell constant`=kappaxxR=(1.4" S "m^(-1))(50Omega)=70m^(-1)`
Case II: `C=0.5" M ",R=280Omega," cell cont."=70m^(-1)`
`kappa=(1)/(R)xx"cell constant"=(1)/(280Omega)xx70m^(-1)`
Molar conductivity (in S `m^(2)mol^(-1)`)
`=(kappa(S" "m^(-1)))/("Molarity"(mol " "L^(-1))xx1000" L "m^(-3))`
`=(0.25" S "m^(-1))/(0.5 " mol "L^(-1)xx1000" L "m^(-3))`
`=5xx10^(-4)" S "m^(2)mol^(-1)`