1.

The molarconductivites wedge_(NaOAc)^(@) and wedge_(HCI)^(W)at infinitedilutionin water at 25^(@)Care 91.0 and 426.2 cm^(2)//"mol"respecitvely Tocalculatewedge_(HoAc)^(@)the additional valuerequiredis

Answer»

`wedge_(H_(2)O)^(@)`
`wedge_(KCI)^(@)`
`wedge_(NaOH)`
`wedge_(NACI)`

SOLUTION :Accordingto KOHLRAUSCH 'slaw
`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)^(@)`thusadditoinal valuerequired is `wedge_(NaCI)^(@)`


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