1.

Calculate the degree of dissociation (alpha) of acetic acid if its molar conductivity is 39.05 S cm^(2)mol^(-1). Given lamda^(@)(H^(+))=349.6S" "cm^(2)mol^(-1) and lamda^(@)(CH_(3)CO O^(-))=40.9" S "cm^(2)mol^(-1).

Answer»

Solution :`ALPHA=(wedge_(m))/(wedge_(m)^(@))`
`wedge_(m)^(@)(CH_(3)COOH)=lamda^(@)(CH_(3)COO^(-))+lamda^(@)(H^(+))=40.9+349.6=390.5" S "cm^(2)mol^(-1)`
`wedge_(m)=39.05" S "cm^(2)mol^(-1)`
`thereforealpha=(39.05)/(390.5)=0.1`


Discussion

No Comment Found

Related InterviewSolutions