The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation
`lambda_m^( C)=lambda_m^(oo)-bsqrtC` where `lambda_(m)^( C)`=molar specific conductance
`lambda_(m)^(oo)`=molar specific conductance at infinite dilution
C=molar concentration
`{:("Molar Concentration of NaCl","Molar Conductance" "In" "ohm"^(-1)cm^(2)"mole"^(-1)),(4xx10^(-4),107),(9xx10^(-4),97),(16xx10^(-4),87):}`
When a certain conductivity cell (C) was filled with `25xx10^(-4)`(M) NaCl solution.The resistance of the cell was found to be 1000 ohm.At Infinite dilution, conductance of `Cl^(-)` and `SO_4^(-2)` are `80 ohm^(-1) cm^(2) "mole"^(-1) and 160 ohm^(-1) cm^2 "mole"^(-1)` respectively.
What is the molar conductance of NaCl at infinite dilution ?

Options

A. `19.25ohm^(-1)cm^(2)"mole"^(-1)`
B. `96.25ohm^(-1)cm^(2)"mole"^(-1)`
C. `385ohm^(-1)cm^(2)"mole"^(-1)`
D. `192.5ohm^(-1)cm^(2)"mole"^(-1)`

Answer

D

Explanation

For `Na_(2)SO_(4)` solution
`K=((l)/(a))xx(1)/(R)=(0.1925)/(400)=4.81xx10^(-4)ohm^(-1)cm^(-1)`
`lamda_(m)=(Kxx1000)/(M)=(4.81xx10^(-4)xx1000)/((5)/(2)xx10^(-3))`
`lamda_(m)(Na_(2)SO_(4))=192.4ohm^(-1)cm^(2)"mole"^(-1)`