The conductivity of 0.001 `" mol "L^(-1)` solution of `CH_(3)COOH` is `4.95xx10^(-5)" S "cm^(-1)`. Calculate its molar conductance and degree of dissociation (alpha).`"Given" lambda_((H^(+)))^(@)=349.6" S "cm^(2)mol^(-1), lambda_((CH_(3)COO^(-)))^(@)=40.95 cm^(2)mol^(-1)`

The conductivity of 0.001 `" mol "L^(-1)` solution of `CH_(3)COOH` is

1.	The conductivity of 0.001 `" mol "L^(-1)` solution of `CH_(3)COOH` is `4.95xx10^(-5)" S "cm^(-1)`. Calculate its molar conductance and degree of dissociation (alpha).`"Given" lambda_((H^(+)))^(@)=349.6" S "cm^(2)mol^(-1), lambda_((CH_(3)COO^(-)))^(@)=40.95 cm^(2)mol^(-1)`
Answer» `k=4.95xx10^(-5)" S "cm^(-1)` `C=0.001" mol "L^(-1)=((0.001" mol"))/((10^(3)"cm"^(3)))=10^(-6)" mol "cm^(-3)` Molar conductance at given concentration `Lambda_(m)^(c)=(k)/(C )=((4.95xx10^(-5)" S "cm^(-1)))/((10^(-6)" mol "cm^(-3)))=49.5" S "cm^(2)mol^(-1)`. Molar conductance at infinite dilution. `(Lambda^(@))` `Lambda_(m)^(@)=(lambda_(CH_(3)COO^(-))^(@)+lambda_(H^(+))^(@))=(40.9+349.6)" S "cm^(2)mol^(-1)` `=390.5" S "cm^(2)mol^(-1)` Degree of dissociation `(alpha)=(Lambda_(m)^(c))/(Lambda_(m^(@)))=((49.5" S "cm^(2)mol^(-1)))/((390.5" S "cm^(2)mol^(-1)))=0.127`

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The conductivity of 0.001 `" mol "L^(-1)` solution of `CH_(3)COOH` is `4.95xx10^(-5)" S "cm^(-1)`. Calculate its molar conductance and degree of dissociation (alpha).`"Given" lambda_((H^(+)))^(@)=349.6" S "cm^(2)mol^(-1), lambda_((CH_(3)COO^(-)))^(@)=40.95 cm^(2)mol^(-1)`

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