0.5 normal solution of a salt placed between two platnum electrodes 2.0 cm. apart and of area of cross section 4.0 sq. cm. has a resistance of 25 ohms. Calculate the equivalent conductivity of solution.

0.5 normal solution of a salt placed between two platnum electrodes 2.

1.	0.5 normal solution of a salt placed between two platnum electrodes 2.0 cm. apart and of area of cross section 4.0 sq. cm. has a resistance of 25 ohms. Calculate the equivalent conductivity of solution.
Answer» Solution :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)`