1.

Conductivity of 0.00241 M acetic acid solution is 7.896xx10^(-5)" S "cm^(-1). Calculate its molar conductivity in this solution. If wedge_(m)^(@) for acetic acid be 390.5 S cm^(2)mol^(-1), what would be its dissociation constant?

Answer»


Solution :`wedge_(m)^(C)=(kappaxx1000)/("Molarity")=(7.896xx10^(-5)" S "cm^(-1)xx1000cm^(3)L^(-1))/(0.00241" mol "L^(-1))=32.76" S "cm^(2)mol^(-1)`
`ALPHA=(32.76)/(390.5)=0.084,K=(calpha^(2))/(1-alpha)=(0.00241xx(0.084)^(2))/(1-0.084)=1.85xx10^(-5)`.


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