Question

Electrolyte	$KCl$	$KN{O}_3$	$HCl$	$NaOAc$	$NaCl$
${\Lambda }^{\infty }{\text{(Scm}}^2{\text{mol}}^{-1})$	$149.9$	$145$	$426.2$	$91$	$126.5$

Calculate ${\Lambda }_{HOAc}^{\infty }$ using appropirate molar conductances of the electrolytes listed above at infinite dilution in $H_{2}O$ at ${25}^{o}C$

• A. $217.5{\text{Scm}}^2{\text{mol}}^{-1}$
• B. $390.7{\text{Scm}}^2{\text{mol}}^{-1}$
• C. $552.7{\text{Scm}}^2{\text{mol}}^{-1}$
• D. $517.2{\text{Scm}}^2{\text{mol}}^{-1}$

Answer 1

The correct option is B $390.7{\text{Scm}}^2{\text{mol}}^{-1}$

Given,
${\Lambda }_{HCl}^{\infty }={\lambda }_{H^{+}}^{\infty }+{\lambda }_{C{l}^{-}}^{\infty }=426.2{\text{Scm}}^2{\text{mol}}^{-1}$
${\Lambda }_{AcONa}^{\infty }={\lambda }_{N{a}^{+}}^{\infty }+{\lambda }_{Ac{O}^{-}}^{\infty }=91.0{\text{Scm}}^2{\text{mol}}^{-1}$
${\Lambda }_{NaCl}^{\infty }={\lambda }_{N{a}^{+}}^{\infty }+{\lambda }_{C{l}^{-}}^{\infty }=126.5{\text{Scm}}^2{\text{mol}}^{-1}$
${\Lambda }_{AcOH}^{\infty }=({\lambda }_{H^{+}}^{\infty }+{\lambda }_{C{l}^{-}}^{\infty })+({\lambda }_{N{a}^{+}}^{\infty }+{\lambda }_{Ac{O}^{-}}^{\infty })-({\lambda }_{N{a}^{+}}^{\infty }+{\lambda }_{C{l}^{-}}^{\infty })=[426.2+91.0-126.5]{\text{Scm}}^2{\text{mol}}^{-1}=390.7{\text{Scm}}^2{\text{mol}}^{-1}$