Van't Hoff factor of $BaCl_2$ in its aqueous solution will be: ($BaCl_2$ is $60 \%$ ionized in the solution)

In an aqueous solution of $NaCl$,the mole fraction of the solvent is $0.9$. Then,the relative lowering in vapour pressure will be ......... $atm$.

When $10^{-3} \ M$ solution of glucose in water freezes at $-0.0186^{\circ} C$,then at what temperature $10^{-3} \ M$ solution of $NaCl$ will freeze (in $^{\circ} C$)?

The measured osmotic pressure of a solution prepared by dissolving $17.4 \ mg$ of $K_2SO_4$ in $2 \ L$ of water at $27^{\circ} C$ is $3.735 \times 10^{-3} \ bar$. The van't Hoff factor is $(R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$; atomic weights $K = 39, S = 32, O = 16)$.

$0.01 \ M$ solutions of $KCl$ and $BaCl_2$ are prepared in water. The freezing point of $KCl$ is found to be $-2 \ ^oC$. What is the freezing point of $BaCl_2$ if it is completely ionised? ............. $^oC$

In comparison to a $0.01 \ M$ solution of glucose,the depression in freezing point of a $0.01 \ M \ MgCl_2$ solution is (Molecular weight of $MgCl_2 = 95$,molecular weight of glucose $= 180$).