‹ Back to the paper
Answer the following.
The molar conductivity of NaCl, CH₃COONa and HCl at infinite dilution is 126·45, 91·0 and 426·16…
The molar conductivity of $\text{NaCl}$, $\text{CH}_3\text{COONa}$ and $\text{HCl}$ at infinite dilution is $126\cdot45$, $91\cdot0$ and $426\cdot16\text{ ohm}^{-1}\text{ cm}^2\text{ mol}^{-1}$ respectively. Calculate the molar conductivity ($\lambda_m^\infty$) for $\text{CH}_3\text{COOH}$ at infinite dilution.
Answer
Answer
Formula used: Kohlrausch's law: Lm0 = L1 + L2 - L3: Lm0 = L1 + L2 - L3 with L1 = 91.0 S cm^2 mol^-1, L2 = 426.16 S cm^2 mol^-1, L3 = 126.45 S cm^2 mol^-1, so Lm0 = 390.71 S cm^2 mol^-1
By Kohlrausch's law:
$\lambda_m^\infty(\mathrm{CH_3COOH}) = \lambda_m^\infty(\mathrm{CH_3COONa}) + \lambda_m^\infty(\mathrm{HCl}) - \lambda_m^\infty(\mathrm{NaCl})$
$= 91\cdot0 + 426\cdot16 - 126\cdot45 = 390\cdot71$
Final answer: 390.71 ohm^-1 cm^2 mol^-1