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.

In a conductivity cell the two platinum electrodes each of area 10 sq.

1.	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» Solution :SINCE the electroces of the CELL are just half dipped, the effective area will be 5 sq cm. Cell constant `= (1)/(a) = (1.5)/(5) = 0.3 cm^(-1)`. Specific conductance = conductance `xx` cell constant `= (1)/("resistance") xx "cell constant"` `= (1)/(50) xx 0.3 = (3)/(500) = "mho cm"^(-1)`. Equivalent conductance = specific conductance `xx` volume `= (3)/(500) xx 20000 = 120 "mho cm"^(2)`. `(0.05 N = N//20 therefore V = 20,000 c c)`