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The molar conductivity of 0.025 mol L^(-...

The molar conductivity of `0.025 mol L^(-1)` methanoic acid is `46.1 S cm^(2) mol^(-1)`. Its degree of dissociation `(alpha)` and dissociation constant. Given `lambda^(@)(H^(+))=349.6 S cm^(-1)` and `lambda^(@)(HCOO^(-)) = 54.6 S cm^(2) mol^(-1)`.

Text Solution

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`Lambda_((HCOOH))^(@) = lambda_((H^(+)))^(@) + lambda_((HCOO^(-)))`
`= 349.6 + 54.6`
`= 404.2 S cm^(2) mol^(-1)`
`alpha = (Lambda_(c)^(m))/(Lambda_(0)^(m)) = (46.1)/(404.2) = 0.114`
`K_(c) = ( c alpha^(2))/(1 - alpha)`
`= (0.025 xx (0.114)^(2))/(1 - 0.114)`
`= 3.67 xx 10^(-4)`
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