Question

The molar conductivities at infinite dilution of
potassium chloride, hydrochloric acid and potas-
sium acetate are 130-1, 379.4 and 95-60-'cm
mor respectively. Calculate the value of molar
conductivity at infinite dilution for acctic acid. If
the molar conductivity of a given acetic acid solu-
tion is 48-5 2-1 cmmorat 25°C, calculate the
degree of dissociation of acetic acid at this
temperature.​

Answer 1

$\huge \: \red \star \: { \large \pink { \underline\mathcal{ANSWER}} } \: \red\star \: </p><p>$

$\\ \green{ \mathrm{concept}} </p><p>$

• kholrausch's law

• $\\ \footnotesize{ \bold{\alpha = \frac{ \land_m }{ { \land_m}^{ o}} }}$

• $\footnotesize{ \text{where} \: \: \: \alpha } \: \: \: \text{is degree of dissociation. }$

$\\ \\ \green{ \mathrm{solution}}$

$\\ {\land_m}^{ o}( \footnotesize{CH_3COOH}) = \\ {\land_m}^{ o}(\footnotesize{CH_3COOK}) + {\land_m}^{ o}(\footnotesize{HCl}) - {\land_m}^{ o}(\footnotesize{KCl}) \\ \\ {\land_m}^{ o}( \footnotesize{CH_3COOH}) =130.1 + 379.4 - 95.60 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 413.90 \: \text{ unit }\: ( \text{same \: as \: given} )$

$\\ \\ \footnotesize{ \bold{\alpha = \frac{ \land_m }{ { \land_m}^{ o}} }} = \frac{48.5}{413.30} = 0.113$

hope it helps