Question

Equivalent conductance of 1M $C{H}_{3}COOH$ is $10oh{m}^{-1}c{m}^{2}equi{v}^{-1}$and that at infinite dilution is $200oh{m}^{-1}c{m}^{2}equi{v}^{-1}$ . Hence the percentage ionization of $C{H}_{3}COOH$ is:

• A. $5$%
• B. $2$%
• C. $4$%
• D. $1$%

Answer 1

The correct option is A $5$%

The degree of the dissociation is equal to the ratio of the molar conduction at dilution v to the molar conduction at infinite dilution. $\alpha =\frac{{\Lambda }_{eq}^v}{{\Lambda }_{eq}^{\infty }}$.

Substitute ${\Lambda }_{eq}^{\infty }=200{ohm}^{-1}{cm}^2{equiv}^{-1}$ and ${\Lambda }_{eq}^v=10{ohm}^{-1}{cm}^2{equiv}^{-1}$ in the above equation.

$\alpha =\frac{10{ohm}^{-1}{cm}^2{equiv}^{-1}}{200{ohm}^{-1}{cm}^2{equiv}^{-1}}=0.05=5$%.

Hence, correct option is $A$.