Question

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.
It 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)`.
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 1

Correct Answer - D
`lambda=(Kxx1000)/(m)`
`=(G^(**)xx1000)/("R m")=(0.1925xx1000)/(400xx[(5xx10^(-3))/(2)])=(192.5xx2)/(2000xx10^(-3))=192.5`