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.
If the cell ( C) is filled with `5xx10^(-3)(N)Na_(2)SO_(4)` the obserbed resistance was 400 ohm.What is the molar conductance of `Na_(2)SO_(4)`.

Options

A. `0.385 cm^(-1)`
B. `3.85 cm^(-1)`
C. `38.5 cm^(-1)`
D. `0.1925 cm^(-1)`

Answer

D

Explanation

For `25xx10^(-4)` (M) NaCl solution
`lambda_m=lambda_m^oo-bsqrtC`
`lambda_m=127-10^3(25xx10^(-4))^(1//2)`
`lambda_m=127-10^(3)xx5xx10^(-2)`
`lambda_m=77`
But `lambda_m=(Kxx1000)/M " " K=(l/a)xx1/R`
`lambda_m=(l/a)xx1/Rxx1000/M`
`lambda_m`=[Cell constant]`xx1000/(RxxM)`
`implies `77=[Cell constant]`xx1000/(1000xx25xx10^(-4))`
Cell constant `=77xx25xx10^(-4)=0.1925 cm^(-1)`