The conductivity of $0.04 \ M \ BaCl_2$ solution is $0.0112 \ \Omega^{-1} \ cm^{-1}$ at $25^{\circ} C$. What is its molar conductivity?

Equivalent conductances of $BaCl_2$,$H_2SO_4$ and $HCl$ at infinite dilution are $\Lambda_1^\infty$,$\Lambda_2^\infty$ and $\Lambda_3^\infty$ respectively. Then,the equivalent conductance of $BaSO_4$ at infinite dilution will be:

The resistivity of a $0.8 \ M$ solution of an electrolyte is $5 \times 10^{-3} \ \Omega \ cm$. Its molar conductivity is $..... \times 10^4 \ \Omega^{-1} \ cm^2 \ mol^{-1}$. (Nearest integer)

Which among the following concentrations of $KCl$ solution is not used to determine the cell constant of a conductivity cell?

Calculate $\Lambda _m^o$ for $CaCl_2$ and $MgSO_4$ from the data given in the table below:

Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$	Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$
$H^{+} : 349.6$	$OH^{-} : 199.1$
$Na^{+} : 50.1$	$Cl^{-} : 76.3$
$K^{+} : 73.5$	$Br^{-} : 78.1$
$Ca^{2+} : 119.0$	$CH_3COO^{-} : 40.9$
$Mg^{2+} : 106.0$	$SO_4^{2-} : 160.0$

The equivalent conductances of two strong electrolytes at infinite dilution in $H_2O$ (where ions move freely through a solution) at $25^{\circ}C$ are given below:
$\lambda^{\infty}_{CH_3COONa} = 91.0 \ S \ cm^2 \ eq^{-1}$
$\lambda^{\infty}_{HCl} = 426.25 \ S \ cm^2 \ eq^{-1}$
What additional information/quantity is needed to calculate $\lambda^{\infty}$ of an aqueous solution of acetic acid?