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 $= 180$).

$17.4\% \text{ (mass/volume) } K_2SO_4$ solution at $27^\circ C$ is isotonic to $5.85\% \text{ (mass/volume) } NaCl$ solution at $27^\circ C$. If $NaCl$ is $100\%$ ionized,what is the $\%$ ionization of $K_2SO_4$ in aqueous solution? [Atomic weights: $K = 39, Na = 23, S = 32, O = 16, Cl = 35.5$]

If $1$ mole of a solute dissociates to give $n$ number of ions,then the degree of dissociation $\alpha$ in terms of the van't Hoff factor $i$ is given by:

Between aqueous solutions of $0.01 \ m \ KCl$ and $0.01 \ m \ BaCl_2$ (strong electrolytes),the freezing point of the $KCl$ solution is $-2^{\circ}C$. The freezing point of the $BaCl_2$ solution will be ..... $^{\circ}C$.

$80$ mole percent of $MgCl_2$ is dissociated in aqueous solution. The vapour pressure of $1.0$ molal aqueous solution of $MgCl_2$ at $38^{\circ} C$ is $.........$ $mm$ $Hg$. (Nearest integer)
Given : Vapour pressure of water at $38^{\circ} C$ is $50$ $mm$ $Hg$.

The freezing point of a $0.2 \ m$ aqueous solution of a weak acid $(HX)$ which is $20\%$ ionized is ......... $^oC$. (Given: $K_f = 1.86 \ ^oC/m$ for water)