The conductivity of 0.001 mol `L^(-1)` solution of `CH_(3)COOH` is `3.905xx10^(-5)S cm^(-1)`. Calculate its molar conductivity and degree of dissociation `(alpha)`. `("Given":lamda_((H^(+)))^(@)=349.65 S cm^(2)mol^(-1)andlamda^(@)(CH_(3)COO^(-))=40.9 D cm^(2)mol^(-1))`

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

1.	The conductivity of 0.001 mol `L^(-1)` solution of `CH_(3)COOH` is `3.905xx10^(-5)S cm^(-1)`. Calculate its molar conductivity and degree of dissociation `(alpha)`. `("Given":lamda_((H^(+)))^(@)=349.65 S cm^(2)mol^(-1)andlamda^(@)(CH_(3)COO^(-))=40.9 D cm^(2)mol^(-1))`
Answer» Correct Answer - `Delta_(m)=39.05 S cm^(2) mol^(-1),a=0.1` Step I. Calcualtion of molar conductance (K) `k=3.905xx10^(5)" S "cm^(-1),C=0.001" mol "L^(-1)=(0.001" mol")/(10^(3) cm^(3))=10^(-6)"mol "cm^(-3)` Molar conductance `(Lambda_(m)) =(k)/(C )=((3.905xx10^(-5)" S "cm^(-1)))/((10^(-6)"mol "cm^(-3)))=39.05" S "cm^(2)mol^(-1)` Step II. Calcualtion of degree of dissociation `(alpha)` `Lambda_(m)^(@)=lamda^(@)(H^(+))+lambda^(@)(CH_(3)COO^(-))` `=(349.6+40.9)=390.5" S "cm^(2)mol^(-1)` `alpha=(Lambda_(m)^(C))/(Lambda_(m)^(o))=((39.05" S "cm^(2)mol^(-1)))/((390.5" S "cm^(2) mol^(-1)))=0.1`

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The conductivity of 0.001 mol `L^(-1)` solution of `CH_(3)COOH` is `3.905xx10^(-5)S cm^(-1)`. Calculate its molar conductivity and degree of dissociation `(alpha)`. `("Given":lamda_((H^(+)))^(@)=349.65 S cm^(2)mol^(-1)andlamda^(@)(CH_(3)COO^(-))=40.9 D cm^(2)mol^(-1))`

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