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 $.0$)

$0.01 \ M$ solutions of glucose,phenol,and potassium chloride were prepared in water. The boiling points of these solutions follow the order:

$17.4\% (W/V)$ of potassium sulphate $(mol. wt. = 174)$ is isotonic with $4\% (W/V)$ aqueous solution of $NaOH$. If $NaOH$ is $100\%$ ionised,the degree of ionisation of potassium sulphate is $.......... \%$.

Of the following four aqueous solutions,the total number of those solutions whose freezing point is lower than that of $0.10 \, M \, C_{2}H_{5}OH$ is (Integer answer).
$(i)$ $0.10 \, M \, Ba_{3}(PO_{4})_{2}$
$(ii)$ $0.10 \, M \, Na_{2}SO_{4}$
$(iii)$ $0.10 \, M \, KCl$
$(iv)$ $0.10 \, M \, Li_{3}PO_{4}$

An aqueous solution of a salt $MX_2$ at a certain temperature has a van't Hoff factor of $2$. The degree of dissociation for this solution of the salt is

For $0.1 \ M$ solutions of the following electrolytes,the colligative property will follow the order: