The molar conductance of NaCl, HCl and `CH_(3)COONa` at infinite dilution are 126.45, 426.16 and 91 `ohm^(-1) cm^(2) mol^(-1)` respectively. The molar conductance of `CH_(3)COOH` at infinite dilution is :

Options

A. `201.28ohm^(-1)cm^(2)mol^(-1)`
B. `390.71ohm^(-1)cm^(2)mol^(-1)`
C. `698.28ohm^(-1)cm^(2)mol^(-1)`
D. `540.48ohm^(-1)cm^(2)mol^(-1)`

Answer

B

Explanation

`(^^)_(m)^(@)(CH_(3)COOH)`
`=(^^)^(@)(CH_(3)COONa)+(^^)^(@)(HCl)-(^^)^(@)(NaCl)`
`=91+426.16-126.45=390.71ohm^(-1)cm^(2)mol^(-1)`