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 $.0$)

In a $0.2 \ molal$ aqueous solution of a weak acid $HX$,the degree of ionisation is $0.25$. The freezing point of the solution will be nearest to ......... $^oC$. $(K_f = 1.86 \ K \ kg \ mol^{-1})$

If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution,the change in freezing point of water $(\Delta T_f),$ when $0.01 \ mol$ of sodium sulphate is dissolved in $1 \ kg$ of water,is $(K_f = 1.86 \ K \ kg \ mol^{-1})$ (in $K$).

Which of the following is a non-electrolyte?

The freezing point depression constant for water is $1.86 \, ^oC \, kg \, mol^{-1}.$ If $5.00 \, g$ of $Na_2SO_4$ is dissolved in $45.0 \, g$ of $H_2O,$ the freezing point is lowered by $3.82 \, ^oC.$ Calculate the van't Hoff factor for $Na_2SO_4.$

The freezing point of a $0.05 \ m \ BaCl_2$ solution in water ($100 \%$ ionisation) is about $(K_f = 1.86 \ K \ kg \ mol^{-1})$. (in $^oC$)