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The conductivity of 0.001 "mol" L^(-1) s...

The conductivity of 0.001 `"mol" L^(-1)` solution of `CH_(3)COOH " is " 3.905 xx 10^(-5) "S" cm^(-1)`. Calculate its molar conductivity and degree of dissociation `(alpha)`.
`"Given" lambda^(@) (H^(+)) = 349.6 "S" cm^(2) "mol"^(-1) " and
" lambda^(0) (CH_(3)COO^(-)) = 40.9 "S" cm^(2) per mol)

Text Solution

Verified by Experts

Using formula, `wedg _(m)^(c ) = (kxx1000)/(c )`
Given `k= 3.905 xx10^(-5) Scm^(-1)` `therefore wedge_(m)^(c)= (3.905xx10^(-5)xx1000)/(0.001)= 39.05cm^(2)mol^(-1)`
`C= 0.001 mol L ^(-1)`
The degree of dissociation,
`alpha = (wedge_m^(c )/(wedge_(m)^(c ))= (39.05)/(390.5) = 0.1` `[ therefore wedge_(m)^(@) = 349.6+ 40 .9 = 390.5 S cm^(2) mol^(-1) ]`
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