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{:("Electrolyte:",KCl,KNO_(3),HCl,NaOAc,NaCl),(wedge^(oo)(Scm^(2)mol^(-1)),149.9,145.0,426.2,91.0,126.5):} Calcualte wedge_(HOAc)^(oo) using appropriate molar conductances of the electrolytes listed above at infinite dilution in H_(2)O at 25^(@)C.

Answer»

517.2
552.7
390.7
217.5

Solution :`wedge_(HOAC)^(OO)=wedge_(NAOAC)^(oo)+wedge_(HCL)^(oo)-wedge_(NaCl)^(oo)` ltBrgt `=91.0+426.2-126.5=390.7`


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