Question

The conductivity of 0.20 mol `L^(-1)` solution of KCI is `2.48xx10^(-2)S cm^(-1)`. Calculate its molar conductivity and degree of dissociation `(alpha)`. Given `lambda_((K^(+)))^(@)=73.5 S cm^(-2)mol^(-1)and lambda_((CI^(-)))^(@)=76.5 mol^(-1)`
`lambda_((K^(+)))^(@)=73.5 S cm^(-2)mol^(-1)and lambda_((CI^(-)))^(@)=76.5 mol^(-1)`

Answer 1

Correct Answer - 0.105
Step I. Calculation of molar conductivity of acetic acid
Molar concentration (C )`=(10^(-3)mol)/(1L)=(10^(-3)mol)/(10^(3)cm^(-3))=10^(-6)" mol cm"^(-3)`
Molar conductivity `(Lambda_(m)^(c ))=(k)/(C )=((4.1xx10^(-5)ohm^(-1)cm^(-1)))/((10^(-6)mol cm^(-3)))=41ohm^(-1)cm^(2)mol^(-1)`
Step II. Calculation of degree of dissociation of acetic acid
`alpha=(Lambda_(m)^(c ))/(Lambda_(m)^(@))=((41" ohm"^(-1)cm^(2)mol^(-1)))/((390.5" ohm"^(-1)cm^(2)mol^(-1)))=0.105`