The resistance of 0.15 M solution of an electrolyte is `50Omega`. The specific conductance of the solution is 2.4 `Sm^(-1)`. The resistance of 0.5 N solution of the same electrolyte measured using the same conductivity cell is `480Omega`. Find the equivalent conductivity of 0.5 N solution of the electrolyte.

Explanation

Given that
`R_(1)=50 Omega" "R_(2)=480Omega`
`kappa_(1)=2.4Sm^(-1) " "kappa_(2)=?`
`N_(1)=0.15N" "N_(2)=0.5N`
`wedge=(kappa_(2)(Sm^(_1)) times 10^(-3)("gram equivalent")^(-1)m^(3))/(N)`
`" "=(0.25 times 10^(-3)S("gram equivalent")^(-1)m^(2))/(0.5)`
`wedge=5 times 10^(-4)Sm^(2)" gram equivalent"^(-1)`
We know that
`kappa="Cell constant"/R`
`therefore kappa_(2)/kappa_(1)=R_(1)/R_(2)`
`kappa_(2)=kappa_(1) times R_(1)/R_(2)`
`" "=2.4Sm^(-1) times (50 Omega)/(480 Omega)`
`" "=0.25Sm^(-1)`