The molar conductivites `wedge_(NaOAc)^(@)` and `wedge_(HCI)^(W)` at infinite dilution in water at `25^(@)C` are 91.0 and 426.2 `cm^(2)//"mol"` respecitvely To calculate `wedge_(HoAc)^(@)` the additional value required is

Options

A. `wedge_(H_(2)O)^(@)`
B. `wedge_(KCI)^(@)`
C. `wedge_(NaOH)`
D. `wedge_(NaCI)`

Answer

C

Explanation

According to kohlrausch 's law
`wedge_(CH_(3)COOH)^(@)=wedge_(CH_(3)COOH_(-))^(@) +wedge_(H^(+))^(@)`
`wedge_(HCI)^(@)=wedge_(H^(+))^(@) +wedge_(CI_(-))^(@)`
To calculate `wedge_("GOAC")^(@) =wedge_("HCI")+wedge_(NaOAC)^(@)-wedge_(NaCI)^(@)` thus additoinal value required is `wedge_(NaCI)^(@)`