Question

Given the following molar conductivity at infinite dilution and ${25}^o$C
HCl: ${\wedge }_m^{\infty }$ = $426.2$ S ${cm}^2$ ${mol}^{-1}$ KCl: ${\wedge }_m^{\infty }$ = 114.42 S ${cm}^2$ ${mol}^{-1}$ $C{H}_{3}COOK$: ${\wedge }_m^{\infty }$ = $149.86$ S ${cm}^2$ ${mol}^{-1}$
The molar conductance at infinite dilution and ${25}^o$C, for acetic acid solution is:

• A. $390.76$ $S{cm}^2$${mol}^{-1}$
• B. $461.86$ $S{cm}^2$${mol}^{-1}$
• C. $289.63$ $S{cm}^2$${mol}^{-1}$
• D. $172.94$ $S{cm}^2$${mol}^{-1}$

Answer 1

Answer: (B)

$461.86$ $S{cm}^2$${mol}^{-1}$

Given :-

$HCl$, ${\Lambda }_m^{\infty }=426.2S{cm}^2{mol}^{-1}$

$KCl$, ${\Lambda }_m^{\infty }=114.42S{cm}^2{mol}^{-1}$

${CH}_{3}COOK,{\Lambda }_m^{\infty }=149.86S{cm}^2{mol}^{-1}$

${\Lambda }_m^{\infty }\left({CH}_{3}COOH\right)={\Lambda }_m^{\infty }\left({CH}_{3}COOK\right)+{\Lambda }_m^{\infty }\left(HCl\right)-{\Lambda }_m^{\infty }\left(KCl\right)$

${\Lambda }_m^{\infty }=461.64S{cm}^2{mol}^{-1}$

$\left({CH}_{3}COOH\right)$