Conductivity of 2.5 x ${{10}^{-4}}$ M methanoic acid

cbse
electrochemistry

#1

Conductivity of 2.5 x ${{10}^{-4}}$ M methanoic acid is 5.25 x ${{10}^{-5}}$ S ${{cm}^{-1}}$. Calculate its molar conductivity and degree of dissociation.


#2

Conductivity of methanoic acid = 5.25 x ${{10}^{-5}}$ S ${{cm}^{-1}}$

Concentration of methanoic acid = 2.5 X ${{10}^{-4}}$ M = 2.5 x ${{10}^{-4}}$ x 1000 mol ${{cm}^{-3}}$ = 0.25 mol ${{cm}^{-3}}$

Molar conductivity,

$A_{ m }$ = K/C = 5.25 x ${{10}^{-5}}$ S ${{cm}^{-1}}$ / 0.25 mol ${{cm}^{-3}}$

= 21 x ${{10}^{-5}}$ S ${{cm}^{2}}$ ${{10}^{-5}}$ S ${{mol}^{-1}}$