The molar conductance of NaCl vauies 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","Molar conductance of NaCl 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 `CI^(-)` 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. `3.85cm^(-1)`
C. `38.5cm^(-1)`
D. `0.1925cm^(-1)`

Answer

D

Explanation

`lambda_(c)=lambda_(oo)-bsqrtc`
`=127-10^(3)sqrt(25xx10^(-4))=77`
`lambda_(c)=(Kxx1000)/(m)`
`77=(Kxx1000)/(25xx10^(-4))`
`K=(77xx25xx10^(-4))/(1000)=77xx25xx10^(-7)`
`K=(G^(**))/(R)`
`G^(**)=KR=77xx25xx10^(-7)xx1000`
`=77xx25xx10^(-4)=0.1925`