The molar conductance of HCl, NaCl and `CH_(3)COONa` are 426,12 6 and 91 `Omega^(-1)cm^(2)"mol"^(-1)` respectively. The molar c onductance for `CH_(3)COOH` is
A. `561Omega^(-1) "mol"^(-1)`
B. `391Omega^(-1)cm^(2)"mol"^(-1)`
C. `261Omega^(-1)cm^(2)"mol"^(-1)`
D. `612Omega^(-1)cm^(2)"mol"^(-1)`

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Correct Answer - B
`^^_(CH_(3)COOH)^(oo)=+^^_( HCl)^(oo)+^^_(CH_(3)COONa^(-))^(oo)-^^_(NaCl)^(oo)`
`=426+91-126=391Omega^(-1)cm^(2)"mol"^(-1)`

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