0.5 N solution of a salt placed between two platinum electrode 2.0cm apart and of area of cross-secton 4.0 sq. cm has a resistance of 25 ohms. Calculate the equivalent conductance of solution.

Explanation

1st step-Calculation of specific conductivity.
here, l=2.0 cm, a=4.0`cm^(2)," "R=25`ohms
`therefore`Conductance`G=(1)/(R)=(1)/(25)ohm^(-1)," Cell constant"=(1)/(a)=(2cm)/(4cm^(2))=(1)/(2)cm^(-1)`
`therefore`Sp. Conductivity (`kappa`)=Observed conductance`xx`cell constant`=(1)/(25)Omega^(-1)xx(1)/(2)cm^(-1)=0.02Omega^(-1)cm^(-1)`
2nd step-Calculation of equivalent conductivity`wedge_(eq)=kappaxx(1000)/(c_(eq))`
Here, `c=0.5N,kappa=0.02ohm^(-1)` (calculated above)
`thereforewedge_(eq)=((0.02Omega^(-1)cm^(-1))xx(1000cm^(3)L^(-1)))/((0.5" g eq "L^(-1)))=40Omega^(-1)cm^(2)eq^(-1)` or `40" S "cm^(2)eq^(-1)`