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The conductivity of 10^(-3) mol/L acetic acid at 25 °C is 4.1 xx 10^(-5) cm^(-1) . Calculate its degree of dissociation,Lambda_(m)^(@) if for acetic acid at 25 °C is 390.5 S cm^(2) mol^(-1). |
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Answer» Solution :Apply the following relation: `Lambda_(m) = (1000 k)/C` Substituting the values in the above EQUATION, we have `Lambda_(m) =(1000 xx 4.1 xx 10^(-5))/10^(-3) = 41 S cm^(2) mol^(-1)` Degree of dissociation is given by the following equation: `alpha = Lambda_(m)^(C )/Lambda_(m)^(@) = 41/(390.5)= 0.105` or `10.5%` |
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