The molar conductivities `Lambda_(NaOAc)^@` and `Lambda_(HCI)^@` at infinite dilution is watter at `25^@C` are `91.0` and `426.2 S cm^@ //`mol respectively. To calculate `Lambda_(HOAc,)^2` the additional value required is:

Options

A. `^^_(H_(2)O)^(o)`
B. `^^_(KCl)^(o)`
C. `^^_(NaOH)^(o)`
D. `^^_(NaCl)^(o)`

Answer

D

Explanation

According to Kohlrausch'slaw the molar conductivity of an electrolyte at infinite dilution is the sum of conductivities of is ions. Therefore to obtain `wedge_(HOAc)^(@)` from the conductivity of `wedge_(NaOAc)^(@)&wedge_(HCl)^(@)` the conductivity value of NaCl i.e., `wedge_(NaCl)^(@)` is also needed.
`wedge_(NaOAc)^(@)+wedge_(HCl)^(@)-wedge_(NaCl)^(@)=wege_(HOAc)^(@)`
i.e., `wedge_(Na^(+))^(@)+wedge_(OAc^(-))^(@)+wedge_(H^(+))^(@)+wedge_(Cl^(-))^(@)-(wedge_(Na^(+))^(@)+wedge_(Cl^(-))^(@))`
`=wedge_(OAc^(-))^(@)+wedge_(H^(+))^(@)=wedge_(HOAc)^(@)`