Applications of Kohlrausch’s Law

Find the solubility product of a saturated solution of ${\text{Ag}}_2{\text{CrO}}_4$  at 298 K in water if the emf of the cell $\text{Ag}\left|{\text{Ag}}^{+}\left({\text{Ag}}_2{\text{CrO}}_4\right)\right|\left|{\text{Ag}}^{+}\left(0\text{.1}\text{M}\right)\right|\text{Ag}\left(\text{s}\right)$ is 0.164 V at 298 K.

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.

The equivalent conductivity of acetic acid at infinite dilution is $387 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{eq}}^{-1}$. At the same temperature, for $0.001 \mathrm{M}$ solution of acetic acid, it is $55 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{eq}}^{-1}$. What is the degree of dissociation of $0.1 \mathrm{N}$ acetic acid? Assume $1-\mathrm{\alpha }\approx 1$

$$\begin{gathered} \mathrm{A} : 1.5\% \\ \mathrm{B} : 1.35\% \\ \mathrm{C} : 2\% \end{gathered}$$

Enter your correct answer as A, B or C.

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 $\mathrm{pH}$ of a $0.1$ molar solution of the acid $\mathrm{HQ}$ is $3$ . The value of the ionisation constant, ${\mathrm{K}}_{\mathrm{a}}$ of this acid is :

The molar conductivities of $\mathrm{KCl}, \mathrm{NaCl}$ and ${\mathrm{KNO}}_3$ at infinite dilution are $152, 128$ and $111 {\mathrm{Scm}}^2 {\mathrm{mol}}^{–1}$ respectively. The molar conductivity of ${\mathrm{NaNO}}_3$ at infinite dilution is :-

The specific conductance of $\mathrm{AgCl}$ solution in water was determined to be $1.8 \times {10}^{–6} {\mathrm{\Omega }}^{–1} {\mathrm{cm}}^{–1} \mathrm{at} 298\mathrm{K}$. The molar conductances at infinite dilution of ${\mathrm{Ag}}^{+}$ & ${\mathrm{Cl}}^{–}$ are $67.9 \& 82.1 {\mathrm{\Omega }}^{-1}1{\mathrm{cm}}^2{\mathrm{mol}}^{–1}$. What is  the solubility of $\mathrm{AgCl}$ in water ?

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 :-