1.

The conductivity of 0.1 M solution of AgNO_3 is 9.47 xx 10^(-3) S cm^(-1) at 291 K . The ionic conductivities of Ag^(+) and NO_(3)^(-) at the same temperature are 55.7 and 50.8 S cm^(2) "equiv"^(-1)respectively . Calculate the degree of dissociation of AgNO_3 in 0.1 M solution .

Answer»


Solution :`Lambda = (KAPPA xx 1000)/(C) = (9.47 xx 10^(-3) xx 1000)/(0.1)`
`= 94.7 S CM^(2) "equiv"^(-1)`
`Lambda^(oo) = lambda^(oo) (Ag^(+)) + lambda^(oo) (NO_(3)^(-)) = 106.5 S cm^(2) "equiv"^(-1)`
`ALPHA = (94.7)/(106.5) = 0.089 or 89 %`


Discussion

No Comment Found

Related InterviewSolutions