The resistance of a `0.01N` solution of an electroyte was found to 210 ohm at `298K` using a conductivity cell with a cell constant of `0.88 cm^(-1)`. Calculate specific conductance and equilvalent conductance of solution.

Explanation

Given, for `0.01N` solution.
`R= 210 ohm`
`(l)/(a) = 0.88 cm^(-1)`
Specific conductance,
`:. k = (1)/(R) xx (l)/(a)`
`k = (1)/(210) xx 0.88 - 4.19 xx 10^(-3) mho cm^(-1)`
`A_(eq) = (k xx 1000)/(N)`
`A_(eq) = (4.19 xx 10^(-3) xx 1000)/(0.01)`
`A_(eq) = 419 mho cm^(2) eq^(-1)`.