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Conductivity of 2.5xx10^(-4) M methanoic...

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

Text Solution

Verified by Experts

`"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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