A potential difference of 20 V applied to the ends of a column of 0.1 M `AgNO_(3)` solution, 4 cm in diameter and 12 cm in length have a current of 20 A. calculate the molar conductance of the solution.

Explanation

Step-I To calculate R.
`V=20V,I=0.2A`
`V=IR implies R=(V)/(I)=(20V)/(0.2A)=100Omega`
`l=12cm,r=4//2cm=2cm`
`R=rho(l)/(a)`
Step II. To calculate
`therefore rho=(Rxxa)/(l)=(Rxxpi r^(2))/(l)`
`=((100Omega)xx(3*142)xx(2)^(2)cm)/(12cm)`
`=(100xx3*142)/(3)Omega" cm"=(3.142)/(3)Omega cm`
`k=(l)/(rho)=(3)/(314*2)Omega^(-1)cm^(-1)`
Step III. To calculate molarity `lamda_(m)=0*1M=0*1mol" "L^(-1)`
`lamda_(m)=(kappaxx1000cm^(3)L^(-1))/("Molarity")`
`=(3)/(314*2)Omega^(-1)cm^(-1)xx(1000cm^(3)L^(-1))/(0*1mol" "L^(-1))`
`=(3000)/(314*2)Omega^(-1)cm^(2)mol^(-1)`
`=95*48Omega^(-1)cm^(2)mol^(-1)`.