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.

Form the following molar conductivities at infinite dilution,

$$\begin{gathered} {\wedge }_{\mathrm{m}}^{\circ } \mathrm{for} {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3=858 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1} \\ {\wedge }_{\mathrm{m}}^{\circ } \mathrm{for} {\mathrm{NH}}_4\mathrm{OH}=238.3 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1} \\ {\wedge }_{\mathrm{m}}^{\circ } \mathrm{for} {\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4=238.4 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1} \end{gathered}$$

Calculate ${\wedge }_{\mathrm{m}}^{\circ }$ for $\mathrm{Al}{\left(\mathrm{OH}\right)}_3$

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 $0.68 {\mathrm{cm}}^{-1}$.

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.

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).

Define limiting molar conductivity $\left(\mathrm{\Lambda }{°}_{\mathrm{m}}\right)$.

What is the molar conductivity of $\mathrm{AgI}$ at zero concentration if the ${\wedge }_0$ values of $\mathrm{NaI}, {\mathrm{AgNO}}_3$ and ${\mathrm{NaNO}}_3$ are respectively $126.9 {\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$, $133.4 {\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$, $121.5 {\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$?

How is the molar conductivity of strong electrolytes at zero concentration determined by graphical method? Why is this method not useful for weak electrolytes?

The molar conductivities at zero concentrations of ${\mathrm{NH}}_4\mathrm{Cl}, \mathrm{NaOH}$ and $\mathrm{NaCl}$ are respectively $149.7{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}, 248.1{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$ and $126.5{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}.$ What is the molar conductivity of ${\mathrm{NH}}_4\mathrm{OH}$ at zero concentration?

What will be the calculated value of standard electrode potential$\left({\mathrm{E}}^0\right) {\mathrm{Cl}}^{-}/\mathrm{AgCl}/\mathrm{Ag}$ at $298 \mathrm{K}$ temperature, if solubility of silver chloride $\left(\mathrm{AgCl}\right)$ in $0.05 \mathrm{M} {\mathrm{BaCl}}_2$ is nearly ${10}^{-9} \mathrm{M} ({\mathrm{E}}^0 \mathrm{Ag}/{\mathrm{Ag}}^{+}=0.80 \mathrm{volts})$.

Which of the following graph is/are correct for a weak acid (HA), where symbols have usual meaning?

The equivalent conductance of $\text{1}\text{M}$ benzoic acid is $\text{12.8 }\mathrm{o}\mathrm{h}{\mathrm{m}}^{-1}\mathrm{c}{\mathrm{m}}^2$ and if the conductance of benzoate ion and ${\mathrm{H}}^{+}$ ion at infinite dilution are $42$ and $\text{288.42}\hspace{0.33em}\mathrm{o}\mathrm{h}{\mathrm{m}}^{-1}\mathrm{c}{\mathrm{m}}^2$ respectively. Calculate percentage of degree of dissociation.
Report your answer by rounding upto nearest whole number.

If molar conductivity of $C{a}^{2+}$ and $C{l}^{-}$ ions are $119$ and $71S c{m}^{2}mo{l}^{-1}$ respectively, then the molar conductivity of $CaC{l}_2$ at infinite dilution is

The graph of $\sqrt{C}\to {\wedge }_m$ for an aqueous solution of which substance is not obtained as a straight line?

Stufy the following cell reaction:

$\mathrm{P}\mathrm{b}\left(\mathrm{s}\right)+{\mathrm{H}\mathrm{g}}_2{\mathrm{S}\mathrm{O}}_4\left(\mathrm{s}\right)\leftrightharpoons \mathrm{ }{\mathrm{P}\mathrm{b}\mathrm{S}\mathrm{O}}_4\left(\mathrm{s}\right)+2\mathrm{H}\mathrm{g}\left(\mathrm{l}\right)$

${\mathrm{E}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^{°}=\mathrm{ }0.92\mathrm{ }\mathrm{V}$

${\mathrm{K}}_{\mathrm{s}\mathrm{p}}\mathrm{ }\mathrm{ }\left({\mathrm{P}\mathrm{b}\mathrm{S}\mathrm{O}}_4\right)=2\mathrm{ }\times \mathrm{ }{10}^{–8},\mathrm{ }{\mathrm{K}}_{\mathrm{s}\mathrm{p}}\mathrm{ }\left({\mathrm{H}\mathrm{g}}_2{\mathrm{S}\mathrm{O}}_4\right)=1\mathrm{ }\times \mathrm{ }{10}^{–6}$

Calculate the ${\mathrm{E}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}$ for a cell containing saturated solutions of these two salts at their respective electrodes:

Given [antilog $\left(0.141\right)=1.38$]

At $298\mathrm{ }\mathrm{K}\mathrm{ }$, the conductivity of a saturated solution of $\mathrm{A}\mathrm{g}\mathrm{C}\mathrm{l}\mathrm{ }$in water is $2.6\mathrm{ }\times \mathrm{ }{10}^{–6}{\mathrm{o}\mathrm{h}\mathrm{m}}^{–1}\mathrm{ }{\mathrm{c}\mathrm{m}}^{–1}\mathrm{ }$ . Given ,${\mathrm{\lambda }}_{\mathrm{m}}^{\mathrm{\infty }}\left({\mathrm{A}\mathrm{g}}^{+}\right)=63\mathrm{ }{\mathrm{o}\mathrm{h}\mathrm{m}}^{–1}\mathrm{ }{\mathrm{c}\mathrm{m}}^2\mathrm{ }{\mathrm{m}\mathrm{o}\mathrm{l}}^{–1}$ & ${\mathrm{\lambda }}_{\mathrm{m}}^{\mathrm{\infty }}\left({\mathrm{C}\mathrm{l}}^{–}\right)=67\mathrm{ }{\mathrm{o}\mathrm{h}\mathrm{m}}^{–1}\mathrm{ }{\mathrm{c}\mathrm{m}}^2\mathrm{ }{\mathrm{m}\mathrm{o}\mathrm{l}}^{–1}$

Therefore solubility product of $\mathrm{A}\mathrm{g}\mathrm{C}\mathrm{l}$ is

The specific conductivity of a saturated solution of AgCl is $\text{3.4}\times 10^{-6}S{\text{cm}}^{-1}$ at 25^oC. If $\lambda \left({\text{Ag}}^{+}\right)=\text{62.3}S{\text{cm}}^2{\text{mol}}^{-1}$, $\lambda \left({\text{Cl}}^{-}\right)=\text{67.7}S{\text{cm}}^2{\text{mol}}^{-1}$, solubility of AgCl at 25^oC is

The conductivity $0.001 \mathrm{M} {\mathrm{Na}}_2{\mathrm{SO}}_4$ solution is $2.6\times {10}^{-4} \mathrm{S} {\mathrm{cm}}^{-1}$ and increases to $7.0\times {10}^{-4} \mathrm{S} {\mathrm{cm}}^{-1}$, when the solution is saturated with ${\mathrm{CaSO}}_4$. The molar conductivities of ${\mathrm{Na}}^{+}$ and ${\mathrm{Ca}}^{+}$ are $50$ and $120 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$, respectively. Neglect conductivity of the water used. What is the solubility product of ${\mathrm{CaSO}}_4$?