Subjective Type

Conductivity of 0.00241 M acetic acid is $$7.896 \times 10^{-5} S \, cm^{-1}$$. Calculate its molar conductivity. If $$\lambda^o_m$$ for acetic acid is $$390.5 S \, cm^2 \, mol^{-1}$$., what is its degree of dissociation ?

Solution

Given:-
Concentration $$\longrightarrow 0.00241M$$
Conductivity (K)= $$7.896\times 10^{-5}S cm^{-1}$$
$$\wedge_m= \cfrac {K}{C}\times 1000$$
$$=\cfrac {7.896}{0.00241}\times 10^{-5}$$
$$\Rightarrow \wedge_m= 0.03276 S cm^{-2} mol^{-1}$$
$$\Rightarrow \wedge^o_m= 390.5 S cm^2 mol^{-1}$$
$$\alpha= \cfrac {\wedge_m}{\wedge^o_m}=\cfrac {0.03276}{390.5}$$
$$\alpha= 8.389\times 10^{-5}$$ .

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