Question

Electrolyte	$KCl$	$KNO_3$	$HCl$	$NaOAc$	$NaCl$
$\Lambda^{\infty} \left(S\,cm^{2}\,mol^{-1}\right)$	$149.9$	$145.0$	$426.2$	$91.0$	$126.5$

Calculate $\Lambda^{\infty}HoAc$ using appropriate molar conductances of the electrolytes listed above at infinite dilution in $H_{2}O$ at $25^{\circ}$ C :

• A. 217.5
• B. 390.7
• C. 552.7
• D. 517.2

Answer 1

Answer: (B)

$ \Lambda^{\infty}_{AcOH} = \Lambda ^{\infty }_{AcON} + \Lambda ^{\infty }_{HCl} - \Lambda ^{\infty }_{NaCl}$
$= 91.0+ 426.2- 126.5$
$= 390.7$