The resistance of a 0.5 M solution of an electrolyte in a conductivity 'cell was found to be 25 ohm. Calculate the molar conductivity of the solution, if the electrodes in the cell are 1.6 cm apart and have an area of `3*2cm^(2)`.

Explanation

Step I : To calculate `kappa`
`R=(1)/(kappa)xx(l)/(a)`
`kappa=(1)/(kappa)xx(l)/(a)`
Here, `R=25Omega,1=1.6cm, and a=3.2cm^(2)`
`therefore kappa=(1)/(25Omega)xx(1.6cm)/(3.2cm^(2))=(1)/(50)Omega^(-1)cm^(-1)`
`=0*02Omega^(-1)cm^(2)`
Step II: To calculate `Lambda_(m)`
`Lambda_(m)=(kappaxx1000cm^(3)L^(-1))/("c (Molarity)")`
Here `kappa=0*02Omega^(-1)cm^(-1)`,
`c=0*5M=0*5mol" "L^(-1)`
`therefore Lambda_(m)=((0*02Omega^(-1)cm^(-1))xx(1000cm^(3)L^(-1)))/((0*05mol" "L^(-1)))`
`=40Omega^(-1)cm^(2)mol^(-1)`