Specific conductivity of N/35 KCl at 298 K is 0.002768 `ohm^(-1) cm^(-1)` and it has resistance of 520 ohm. A N/25 solution of a salt kept in the same cell was found to have a resistance of 300 ohm at 298 K. Calculate equivalent conductance of the solution.

Answer

`119.5 S cm^(2) "equiv"^(-1)`

Explanation

Cell constant = `kappa xx R`
`= 0.002768 xx 520 = 1.43936 cm^(-1)` specific conductivity for salt solution , `kappa ` = Cell constant `xx` conductance
`= 1.43936 xx (1)/(300)`
`= 0.00478 ohm^(-1) cm^(-1)`
`Lambda = (0.00478 xx 1000)/(1//25)`
`= 119.5 S cm^(2) "equiv"^(-1)`