Question

Calculate the molar conductivity of acetic acid at infinite dilution. Given that molar conductivity of $HCl, CH_3COONa$ and $NaCl$ is 426.1. 91.0 and 126.5 $ohm^{-1} cm^2 mol^{-1}$ respectively

• A. $390.6 \,ohm^{-1}\, cm^2\, mol^{-1}$
• B. $195.3\, ohm^{-1}\, cm^2\, mol^{-1}$
• C. $585.9\, ohm^{-1}\, cm^2\, mol^{-1}$
• D. $292.95\, ohm^{-1}\, cm^2\, mol^{-1}$

Answer 1

Answer: (A)

$426.1 = \lambda^{0}_{\left(H^{+}\right)} + \lambda^{0}_{\left(Cl^{-}\right)} ...\left(i\right) $
$91.0 = \lambda^{0}_{\left(Na^{+}\right)} +\lambda^{0}_{\left(CH_3COO^{-}\right)} ...\left(ii\right)$
$ 126.5 = \lambda^{0}_{\left(Na^{+}\right)} + \lambda^{0}_{\left(Cl^{-}\right)} ....\left(iii\right)$
Adding eqn. $(i)$ and $(ii)$ and subtracting eqn. $(iii)$, we get
$\lambda^{0}_{\left(H^{+}\right)} + \lambda^{0}_{\left(CH_3COO^{-}\right)} = 426.1 + 91.0 - 126.5$
$= 390.6 \,ohm^{-1} cm^{2} mol^{-1}$