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

Molar conductivity of a weak acid $HQ$ of concentration $0.18 \ M$ was found to be $1/30$ of the molar conductivity of another weak acid $HZ$ with concentration of $0.02 \ M$. If $\lambda_{Q^{-}}^0 = \lambda_{Z^{-}}^0$, then the difference of the $pK_a$ values of the two weak acids $(pK_a(HQ) - pK_a(HZ))$ is . . . . . . (Nearest integer).
[Given: degree of dissociation $(\alpha)$ $\ll 1$ for both weak acids, $\lambda^0$: limiting molar conductivity of ions]

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............

Which of the following expressions correctly represents the equivalent conductance at infinite dilution of $Al_2(SO_4)_3$? Given that $\Lambda_{Al^{3+}}^o$ and $\Lambda_{SO_4^{2-}}^o$ are the equivalent conductances at infinite dilution of the respective ions.

What must be the molarity of $BaCl_2$ solution to have molar conductivity $240 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$ and conductivity $0.012 \text{ } \Omega^{-1} \text{cm}^{-1}$ at $25^{\circ} \text{C}$ (in $\text{ M}$)?

At $25^o C$,the molar conductivity of a $0.1 \, N \, CH_3COOH$ solution is $80 \, S \, cm^2 \, mol^{-1}$ and the molar conductivity at infinite dilution is $400 \, S \, cm^2 \, mol^{-1}$. What is the degree of dissociation $(\alpha)$ of $CH_3COOH$?