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 $.0$)

The freezing points of equimolar solutions of glucose,$KNO_3$ and $AlCl_3$ are in the order of:

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

For a $0.1 \, molal$ aqueous solution of $2-$iodopropanoic acid,the degree of dissociation is $5 \%$. The freezing point of the solution will be ............ $^o C$.

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).
$(i)$ $0.10 \, M \, Ba_{3}(PO_{4})_{2}$
$(ii)$ $0.10 \, M \, Na_{2}SO_{4}$
$(iii)$ $0.10 \, M \, KCl$
$(iv)$ $0.10 \, M \, Li_{3}PO_{4}$

$0.2 \ molal$ aqueous solution of an electrolyte $A_3B$ is $90\%$ ionised. The boiling point of the solution at $1 \ atm$ is ........ $K.$ ($K_b$ for $H_2O = 0.52 \ K \ Kg \ mol^{-1}$ and $B.pt.$ of $H_2O = 373 \ K$)