K L Chugh Solutions for Chapter: Electrochemistry, Exercise 13: Follow Up Problems

The resistance of $0.01 \mathrm{M} {\mathrm{CH}}_3\mathrm{COOH}$ solution was found to be $2220$ ohm in a conductivity cell having cell constant $0.366{ \mathrm{cm}}^{-1}.$ Calculate:

molar conductivity $\left({\wedge }_{\mathrm{m}}\right)$ of $0.01 \mathrm{M} {\mathrm{CH}}_3\mathrm{COOH}$

$\left[{\mathrm{\lambda }}^0\left({\mathrm{H}}^{+}\right)=349.1 {\mathrm{ohm}}^{-1}{ \mathrm{cm}}^2{ \mathrm{mol}}^{-1}; {\mathrm{\lambda }}^0\left({\mathrm{CH}}_3{\mathrm{COO}}^{-}\right)=40.9{\mathrm{ohm}}^{-1}{ \mathrm{cm}}^2{ \mathrm{mol}}^{-1}\right]$

The resistance of $0.01 \mathrm{M} {\mathrm{CH}}_3\mathrm{COOH}$ solution was found to be $2220$ ohm in a conductivity cell having cell constant $0.366{ \mathrm{cm}}^{-1}.$ Calculate:

${\wedge }_{\mathrm{m}}^{\infty }$

$\left[{\mathrm{\lambda }}^0\left({\mathrm{H}}^{+}\right)=349.1 {\mathrm{ohm}}^{-1}{ \mathrm{cm}}^2{ \mathrm{mol}}^{-1}; {\mathrm{\lambda }}^0\left({\mathrm{CH}}_3{\mathrm{COO}}^{-}\right)=40.9{\mathrm{ohm}}^{-1}{ \mathrm{cm}}^2{ \mathrm{mol}}^{-1}\right]$

The resistance of $0.01 \mathrm{M} {\mathrm{CH}}_3\mathrm{COOH}$ solution was found to be $2220$ ohm in a conductivity cell having cell constant $0.366{ \mathrm{cm}}^{-1}.$ Calculate:

degree of dissociation, $\mathrm{\alpha }$

$\left[{\mathrm{\lambda }}^0\left({\mathrm{H}}^{+}\right)=349.1 {\mathrm{ohm}}^{-1}{ \mathrm{cm}}^2{ \mathrm{mol}}^{-1}; {\mathrm{\lambda }}^0\left({\mathrm{CH}}_3{\mathrm{COO}}^{-}\right)=40.9{\mathrm{ohm}}^{-1}{ \mathrm{cm}}^2{ \mathrm{mol}}^{-1}\right]$

The resistance of $0.01 \mathrm{M} {\mathrm{CH}}_3\mathrm{COOH}$ solution was found to be $2220$ ohm in a conductivity cell having cell constant $0.366{ \mathrm{cm}}^{-1}.$ Calculate:

dissociation constant of the acid

$\left[{\mathrm{\lambda }}^0\left({\mathrm{H}}^{+}\right)=349.1 {\mathrm{ohm}}^{-1}{ \mathrm{cm}}^2{ \mathrm{mol}}^{-1}; {\mathrm{\lambda }}^0\left({\mathrm{CH}}_3{\mathrm{COO}}^{-}\right)=40.9{\mathrm{ohm}}^{-1}{ \mathrm{cm}}^2{ \mathrm{mol}}^{-1}\right]$

The molar conductivity of ${\mathrm{NH}}_4\mathrm{Cl}$ at infinite dilution is $149.7 {\mathrm{ohm}}^{-1}{ \mathrm{cm}}^2{ \mathrm{mol}}^{-1}$ and ionic conductances of hydroxyl ions and chloride ions are $198$ and $76.3$ respectively. Calculate the molar conductivity of ${\mathrm{NH}}_4\mathrm{OH}$ at infinite dilution.

Calculate ${\wedge }^0$ for ${\mathrm{CaCl}}_2$ and ${\mathrm{MgSO}}_4$ from the following data:

${\mathrm{\lambda }}_{{\mathrm{Ca}}^{2+}}^0=119.0 \mathrm{S}{ \mathrm{cm}}^2{ \mathrm{mol}}^{-1}; {\mathrm{\lambda }}_{{\mathrm{Mg}}^{2+}}^0=106 \mathrm{S} {\mathrm{cm}}^2{ \mathrm{mol}}^{-1}$

${\mathrm{\lambda }}_{{\mathrm{Cl}}^{-}}^0=76.3 \mathrm{S}{ \mathrm{cm}}^2{ \mathrm{mol}}^{-1}; {\mathrm{\lambda }}_{{\mathrm{SO}}_4^{2-}}^0=160.0 \mathrm{S}{ \mathrm{cm}}^2{ \mathrm{mol}}^{-1}$

The conductivity of $0.001028 \mathrm{M}$ acetic acid is $4.95\times {10}^{-5} \mathrm{S} {\mathrm{cm}}^{-1}.$ Calculate its dissociation constant if ${\wedge }^0$ for acetic acid is $390.5 \mathrm{S}{ \mathrm{cm}}^2{ \mathrm{mol}}^{-1}.$

Molar conductance of $1.5 \mathrm{M}$ solution of an electrolyte is found to be $138.9 \mathrm{S}{ \mathrm{cm}}^2.$ What would be the specific conductance for this solution.