Kohlrausch's Law and its Application

Calculate ${\Lambda }^{\text{o}}{}_{\text{HOAC}}$  at infinite dilution in  ${\text{H}}_{\text{2}}\text{O}$ at  $\text{25}°\text{C}$:

The molar conductance at infinite dilution for sodium acetate, hydrochloric acid and sodium chloride are $91.0$, $426.2$ and $126.5 \mathrm{S} {\mathrm{cm}}^2{\mathrm{mol}}^{-1}$ respectively at $298 \mathrm{K}$. The molar conductance of acetic acid at infinite dilution would be

The molar conductance and the specific conductance of an electrolyte are related by the formula

The molar conductance of $0.01 \mathrm{M}$ electrolyte is $124 {\mathrm{ohm}}^{-1} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$ at $298 \mathrm{K}$. Calculate its specific conductance.

The specific conductance of $0.01 \mathrm{M}$ of an electrolyte is $1.24\times {10}^{-3} {\mathrm{ohm}}^{-1} {\mathrm{cm}}^{-1}$ at $298 \mathrm{K}$. Calculate the molar conductance.

Calculate the molar conductance of $0.02 \mathrm{M}$ solution of an electrolyte which has a resistance of $310 \mathrm{ohm}$ at $298 \mathrm{K}$. Cell constant is $0.68 {\mathrm{cm}}^{-1}$.

$0.05 \mathrm{M}$ $\mathrm{NaOH}$ offered a resistance of $31.2 \mathrm{ohm}$ in a conductivity cell having a cell constant of $0.38 {\mathrm{cm}}^{-1}$. Calculate the molar conductivity of $\mathrm{NaOH}$ solution.

Calculate the molar conductance of $0.01 \mathrm{M}$ solution of an electrolyte which has a resistance of $210 \mathrm{ohm}$ at $298 \mathrm{K}$. Cell constant is $0.88 {\mathrm{cm}}^{-1}$.

Specific conductance of $0.1 \mathrm{M}$ ${\mathrm{CH}}_3\mathrm{COOH}$ is $4.7\times {10}^{-2} {\mathrm{ohm}}^{-1} {\mathrm{cm}}^{-1}$ at $298 \mathrm{K}$. Calculate its molar conductance.

Which statement is not correct for Kohlrausch law?

If the value of molar conductivities at infinite dilution for ${\mathrm{CH}}_3\mathrm{COOH}$, $\mathrm{HCl}$ and $\mathrm{NaCl}$ are $390.5, 425.4$and $126.4 \mathrm{S} {\mathrm{cm}}^2{\mathrm{mol}}^{-1}$ respectively at $298 \mathrm{K}$. Calculate the molar conductivity at infinite dilution of ${\mathrm{CH}}_3\mathrm{COONa}$ in $\mathrm{S} {\mathrm{cm}}^2{\mathrm{mol}}^{-1}$. (Just mention the value, no requirement of units in the answer).

The molar conductivity of a solution of a weak acid $\mathrm{HX} (0.01 \mathrm{M})$ is $10$ times smaller than the molar conductivity of a solution of a weak acid $\mathrm{HY} (0.10 \mathrm{M})$. If $\mathrm{\lambda }{°}_{{\mathrm{X}}^{-}}=\mathrm{\lambda }{°}_{{\mathrm{Y}}^{-}}$, the difference in their $\mathrm{pKa}$ values, ${\mathrm{pK}}_{\mathrm{a}}\left(\mathrm{HX}\right)-{\mathrm{pK}}_{\mathrm{a}}\left(\mathrm{HY}\right)$, is (consider degree of ionization of both acids to be $<<1$)

${{\mathrm{\Lambda }}^{°}}_{\mathrm{m}}$ for $\mathrm{NaCl}, \mathrm{HCl}$ and $\mathrm{NaAc}$ are $126.4, 425.9$ and $91.0 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$ respectively. Calculate ${{\mathrm{\Lambda }}^{°}}_{\mathrm{m}}$ for $\mathrm{HAc}$.

The maximum molar conductivities of ${\mathrm{X}}^{+3}$ and ${\mathrm{Y}}^{-}$ ions are $200$ and $100 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$ respectively. The molar conductivity of ${\mathrm{XY}}_3$ at infinite dilution is :-

What is meant by limiting molar conductance ?

The molar conductivity of $0.025 \mathrm{mol} {\mathrm{L}}^{-1}$ methanoic acid is $46.1 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$. Calculate the degree of dissociation. Given : ${{\mathrm{\lambda }}^{°}}_{\left({\mathrm{H}}^{+}\right)}=349.6 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$; ${{\mathrm{\lambda }}^{°}}_{\left({\mathrm{HCOO}}^{-}\right)}=54.6 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$.

The molar conductivities at infinite dilution for ${\mathrm{CH}}_3\mathrm{COONa}, \mathrm{HCl}$ and $\mathrm{NaCl}$ are $91.0, 425.9$ and $126.4 \mathrm{S} {\mathrm{cm}}^{2 }{\mathrm{mol}}^{-1}$ respectively at $298$$\mathrm{K}$. Calculate the molar conductivities of ${\mathrm{CH}}_3\mathrm{COOH}$ at infinite dilution.

State Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution?

Consider the galvanic cell, $\mathrm{Pt}\left(\mathrm{s}\right)\overline{){\mathrm{H}}_2(1 \mathrm{bar})\overline{)\mathrm{HCl}\left(\mathrm{aq}\right)(1 \mathrm{M})\overline{)\mathrm{Cl}(1 \mathrm{bar})\overline{)\mathrm{Pt}\left(\mathrm{s}\right)}}}}$. After running the cell for sometime, the concentration of the electrolyte is automatically raised to $3 \mathrm{M} \mathrm{HCl}$. Molar conductivity of the $3 \mathrm{M} \mathrm{HCl}$ is about $240 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$ and limiting molar conductivity of $\mathrm{HCl}$ is about $420 \mathrm{S} {\mathrm{cm}}^2{\mathrm{mol}}^{-1}$. If ${\mathrm{K}}_{\mathrm{b}}$ of water is $0.52 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}$, calculate the boiling point of the electrolyte at the end of the experiment.