In a conductivity cell the two platinum electrodes each of area 10 sq. cm are fixed 1.5 cm apart. The cell. Contained `0.05N` solution of a salt. If the two electrodes are just half dipped into the solution which has a resistance of 50 ohms, find equivalent conductance of the salt solution.

Answer

`120 mho cm^(2) eq^(-1)`

Explanation

Since the electrodes of the cell are just half depped the effective area will be 5 eq cm.
Cell constant `= (1)/(a) =(1.5)/(5) = 0.3 cm^(-1)`
Specific conductance = conductance `x` cell constant
`= (1)/("resistance") xx` cell constant `=(1)/(50)xx0.3 =(3)/(500)"mho cm"^(-1)`
Equivalent conductance = specific conductance `x` volume....(8)
`(3)/(500)xx 20000 = 120 "mho cm"^(2)`
`(0.05 N =N//20 :. V =20,000 "cc")`.