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Conductivity of 2.5xx10^(-4) M methanoic acid is 5.25xx10^(-5)S com^(-1). Calculate its molar conductivity and degreeof dissociation. "Given ":lamda^(0)(H^(+))=349.5 S cm^(2) mol^(-1) and lamda^(0)(HCOO^(-))=50.5 S cm^(2) mol^(-1).

Answer»

Solution :`"Given, "K=5.25xx10^(-5) S CM^(-1)`
`C=2.5xx10^(-4)ML^(-1)`
Then molar conductivity, `^^_(m)=(K)/(C)`
`=(5.25xx10^(-5)S cm ^(-1))/(2.5xx10^(-4)ML^(-1))xx(1000 cm^(3))/(L)=210 cm^(2) M^(-1)`
`OVERSET(@)^^_(m)(HCOOH)=lamda^(@)(H^(+))+lamda^(@)(HCOO^(-))`
`=349.5 S cm^(2) mol^(-1)+50.5 S cm^(2) mol^(-1)=400 S cm^(2) mol^(-1)`
`"Now, "alpha=(^^_(m))/(overset(@)^^_(m))=(210)/(400)=0.525`


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