Question

Conductivity of 0.00241 M acetic acid is $7.896\times {10}^{-5}Sc{m}^{-1}$. Calculate its molar conductivity. If ${\lambda }_m^o$ for acetic acid is $390.5Sc{m}^{2}mo{l}^{-1}$., what is its degree of dissociation ?

Answer 1

Given:-

Concentration $⟶0.00241M$

Conductivity (K)= $7.896\times {10}^{-5}Sc{m}^{-1}$

${\wedge }_m=\frac{K}{C}\times 1000$

$=\frac{7.896}{0.00241}\times {10}^{-5}$

$\Rightarrow {\wedge }_m=0.03276Sc{m}^{-2}mo{l}^{-1}$

$\Rightarrow {\wedge }_m^o=390.5Sc{m}^{2}mo{l}^{-1}$

$\alpha =\frac{{\wedge }_m}{{\wedge }_m^o}=\frac{0.03276}{390.5}$

$\alpha =8.389\times {10}^{-5}$ .