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 cell constant of the conductivity cell (C )

Options

A. `0.385cm^(-1)`
B. `3.86cm^(-1)`
C. `38.5cm^(-1)`
D. `0.1925cm^(-1)`

Answer

D

Explanation

For `25xx10^(-4)` (M) NaCl solution ltbr `lamda_(m)=lamda_(m)^(infty)-bsqrt(C)`
`lamda_(m)=127-10^(3)(25xx10^(-4))^(1//2)`
`lamda_(m)=127-10^(3)xx5xx10^(-2)`
`lamda_(m)=77`
but `lamda_(m)=(Kxx1000)/(M)`:`K=((l)/(a))xx(1)/(R))`
`lamda_(m)=((l)/(a))xx(1)/(R)xx(1000)/(M)`
`lamda_(m)=`[cell constant] `xx(1000)/(RxxM)implies77=`[cell constanta]`xx(1000)/(1000xx25xx10^(-4))`
Cell constant `=77xx25xx10^(-4)=0.1925cm^(-1)`