Given the limiting molar conductivities at $25^{\circ}C$ for the following electrolytes in $S \, cm^{2} \, mol^{-1}$:
$KCl = 149.9$
$KNO_{3} = 145.0$
$HCl = 426.2$
$NaOAc = 91.0$
$NaCl = 126.5$
Calculate $\Lambda^{\infty}_{HOAc}$ using the Kohlrausch law of independent migration of ions.

Given below is the plot of the molar conductivity vs $\sqrt{concentration}$ for $KCl$ in aqueous solution. If,for the higher concentration of $KCl$ solution,the resistance of the conductivity cell is $100 \ \Omega$,then the resistance of the same cell with the dilute solution is '$x$' $\Omega$. The value of $x$ is $............$ ($Nearest$ $integer$)

The values of conductivity of some materials at $298.15 \ K$ in $S \ m^{-1}$ are $2.1 \times 10^3$,$1.0 \times 10^{-16}$,$1.2 \times 10$,$3.91$,$1.5 \times 10^{-2}$,$1 \times 10^{-1}$,$1.0 \times 10^3$. The number of conductors among the materials is............

For an aqueous solution of $CaCl_2$ electrolyte,the graph between molar conductivity $(\lambda_m)$ and $(\text{concentration})^{1/2}$ is:

The molar conductivity of $0.4 \ M \ KCl$ solution is $2.5 \times 10^2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$. What is the resistivity of the solution?

$A$ conductivity cell containing $5 \times 10^{-4} \ M \ NaCl$ solution develops a resistance of $14000 \ \Omega$ at $25^{\circ} C$. Calculate the conductivity of the solution if the cell constant is $0.84 \ cm^{-1}$.