The equivalent conductances at infinite dilution of `HCl, CH_(3)COONa " and " NaCl` are 426.16, 91.0 and 126.45 `ohm^(-1)cm^(2)` gm `"equivalent"^(-1)` respectively. Calculate the equivalent conductance `(lambda_(infty))` of acetic acid.

Explanation

Given :
`lambda_(infty)CH_(3)COONa=91.0ohm^(-1)cm^(2)g" "eq^(-1)`
Formula :
`lambda_(infty)CH_(3)COOH=lambda_(infty)CH_(3)COONa+lambda_(infty)HCl-lambda_(infty)NaCl`
`lambda_(infty)HCl=246.16ohm^(-1)cm^(2)g" "eq^(-1)`
`lambda_(infty)NaCl=126.45ohm^(-1)cm^(2)g" "eq^(-1)`
Solution : `therefore lambda_(infty)CH_(3)COOH`
`" "=91.0+426.16-(126.45)`
`" "=517.16-126.45`
`lambda_(infty)CH_(3)COOH=390.71ohm^(-1)cm^(2)g" "eq^(-1)`