Question

The conductivity of 0.1M solution of AgN03 is 9.47×10^-3scm^-1 at 291k.The ionic conductivity of Ag+ and NO3- at the same temperature are 55.7 and 50.8scm2 equivalent^-1 respectively. Calculate degree of dissociation of AgNO3 in o.1M solution.

Answer 1

Answer:

given = conductivity of 0.1M

            solution of AgN03 is 9.47×10^-3scm^-1

            temprature =  291K

            molar conductivity of

           Ag+ = λ⁰= 55.7cm²

           NO3⁻ =  λ∞ = = 50.8 S cm²

to find =  degree of dissociation of AgNO3 in o.1M solution.

solution  =

             molar conductance of 0.1 M AgNO₃

             λ⁰ = K X 1000/ M

             λ⁰=  9.47 X 10⁻³X 1000/ 0.1

             λ⁰ =  94.7 S cm²mol⁻¹

apply tyhe koharausch law

 λ∞(AgNO₃) =  λ∞(Ag₊) + λ⁰(NO₃⁻)

                     =  55.7 + 50.8

                     =  106.5S cm²mol⁻¹

now for dissociation constant  

                 ∝ = λ⁰/ λ∞

                 ∝ = 94.7 / 106.5 = 0.889

so we can conclude that the the degree of dissociation of AgNO3 in o.1M solution is 0.889