$A$ $0.5\,M\,NaOH$ solution offers a resistance of $31.6\,\Omega$ in a conductivity cell at room temperature. What shall be the approximate molar conductance of this $NaOH$ solution if the cell constant of the cell is $0.367\,cm^{-1}$? (in $S\,cm^2\,mol^{-1}$)

The dissociation constant of acetic acid is $1.6 \times 10^{-5}$ and molar conductance at infinite dilution is $380 \times 10^{-4} \, S \, m^2 \, mol^{-1}$. The specific conductance of $0.01 \, M$ acid solution is

At $291 \ K$,a saturated solution of $BaSO_4$ was found to have a specific conductivity of $3.648 \times 10^{-6} \ \Omega^{-1} \ cm^{-1}$ and that of the water used is $1.25 \times 10^{-6} \ \Omega^{-1} \ cm^{-1}$. If the ionic conductances of $Ba^{2+}$ and $SO_4^{2-}$ are $110$ and $136.6 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively,the solubility of $BaSO_4$ at $291 \ K$ will be (Atomic masses: $Ba=137, S=32, O=16$)

What is the common unit of conductivity if the dimensions are expressed in centimeter?

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

What is the relation between the concentration of solution and molar conductivity $(\Lambda_m)$?