Question

At 291 K the moler conductace values at infinie dilution of `NH_(4)CI`, NaOH and NaCI are 129.1, 217.4 are 108.3 S `cm^(2) mol^(-1)` respectively. Calculate the molar conductane of `NH_(4)OH` at infinite dilution.

Answer 1

Correct Answer - `238.2 S cm^(2) mol^(-1)`
`Lambda_(m(NH_(4)OH))^(oo)=Lambda_(m(NH_(4)Cl))^(oo)+Lambda_(m(NaOH))^(oo)-Lambda_(m(NaCl))^(oo)`
`=(129.1+217.4-108.3)" S "cm^(2)mol^(-1)=238.2" S "cm^(2)mol^(-1)`.