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Conducitivity of 2.5 xx 10^(-4) M methon...

Conducitivity of `2.5 xx 10^(-4)` M methonic acid is `5.25 xx 10^(-5) S cm^(-1)`. Calculate the molar conductivity and degree of dissociation.
Given: `lambda^(@) (H^(+)) = 349.5 S cm^(2) mol^(-1)` and `lambda^(@) (HCOO^(-)) = 50.5 S cm^(2) mol^(-1)`

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`Lambda_(HCOOH)^(@)= Lambda_(H^(+))^(@) + Lambda_(HCOO^(-))^(@) =349.5 + 50.5 = 400 S cm^(2) mol^(-1)`
`Lambda_(HCOOH) = K xx 1000 cm^(3) L^(-1) "molarity"^(-1)`
`=(5.25 xx 10^(-5) S cm^(-1) xx 1000 cm^(3) L^(-1))/(2.5 xx 10^(-4)"mol" L^(-1)) = 210 S cm^(2) mol^(-1)`
Degree of dissociation `(alpha) = Lambda_(m)/Lambda_(,)^(@) = 210/400 = 0.525` or `52.5%`
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