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 $= 180$).

What is the percentage of dissociation of a $0.011 \ m$ aqueous solution of $K_3[Fe(CN)_6]$ having a freezing point of $-0.063 \ ^oC$? (Given: $K_f = 1.86 \ K \ kg \ mol^{-1}$ for water)

$A$ $0.5 \ m$ aqueous solution of a weak acid $(HX)$ is $20\%$ ionized. If $K_f$ for the solvent is $1.86 \ K \ kg \ mol^{-1}$,the depression in freezing point of the solution is ..... $K$.

Calculate the amount of $KCl$ $(M_w = 74.5 \ g \ mol^{-1})$ in grams required to depress the freezing point of $1000 \ g$ of water by $2 \ K$. (Given: $K_f = 1.86 \ K \ kg \ mol^{-1}$) (in $.0$)

$80$ 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 an equimolal aqueous solution will be highest for: