A conductivity cell is used to measure the conductance of electrolyte . It makes use of conductivity of water which does not contain any ions. The cell constant of conductivity cell is determined.
Calculate `alpha` of `CH_(3)COOH` if `Delta^(oo)._(m)` for `HCl, NaCl, CH_(3)COOHNa` are `426,126,91 S cm^(2) mol^(-1)`, respectively, and `wedge_(m)=14.4 S cm^(2) mol^(-1) ` at `0.015 M` concentration.

Options

A. `0.037`
B. `0.018`
C. `0.37`
D. `0.18`

Answer

a

Explanation

`wedge_(m(CH_(3)COOH))^(@)=(426)-126+91`
`=391 Scm^(2) mol^(-1)`
`implies alpha=(14.4)/(391)~~0.037`