Question

Calculate the molar conductivity of acetic acid at infinite dilution. Given that molar conductivity of $HCl,C{H}_{3}COONa$ and $NaCl$ is 426.1. 91.0 and 126.5 $oh{m}^{-1}c{m}^{2}mo{l}^{-1}$ respectively

• A. $390.6oh{m}^{-1}c{m}^{2}mo{l}^{-1}$
• B. $195.3oh{m}^{-1}c{m}^{2}mo{l}^{-1}$
• C. $585.9oh{m}^{-1}c{m}^{2}mo{l}^{-1}$
• D. $292.95oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

Answer 1

Answer: (A)

$426.1={\lambda }_{\left(H^{+}\right)}^0+{\lambda }_{\left(C{l}^{-}\right)}^0...\left(i\right)$
$91.0={\lambda }_{\left(N{a}^{+}\right)}^0+{\lambda }_{\left(C{H}_{3}CO{O}^{-}\right)}^0...\left(ii\right)$
$126.5={\lambda }_{\left(N{a}^{+}\right)}^0+{\lambda }_{\left(C{l}^{-}\right)}^0....\left(iii\right)$
Adding eqn. $(i)$ and $(ii)$ and subtracting eqn. $(iii)$, we get
${\lambda }_{\left(H^{+}\right)}^0+{\lambda }_{\left(C{H}_{3}CO{O}^{-}\right)}^0=426.1+91.0-126.5$
$=390.6oh{m}^{-1}c{m}^{2}mo{l}^{-1}$