Equivalent conductance of 1M `CH_(3)COOH` is `10ohm^(-1) cm^(2) "equiv"^(-1)` and that at inifinite dilution is 200 `ohm^(-1) cm^(2) "equiv"^(-1)`. Hence, % ionisation of `CH_(3)COOJ` is:

Options

A. `5 %`
B. `2 %`
C. `4 %`
D. `1 %`

Answer

A

Explanation

`Lambda_(eq)^(infty)=200" ohm"^(-1)cm^(2)eq^(-1) lambda_(eq)^(v)=10" ohm"^(-1)cm^(2)eq^(-1)`
degree of Ionization `(alpha)=(Lambda_(eq)^(v))/Lambda_(eq)^(v)=10/200=0.05`
.: % of ionization `=0.05xx100=5%`