The freezing point of a $0.05 \ molal$ solution of a non-electrolyte in water is $.......... \ ^oC$. $(K_f = 1.86 \ K \ kg \ mol^{-1})$

The freezing point of an aqueous solution containing $25 \ g$ of ethanol $(C_2H_5OH)$ in $1000 \ g$ of $H_2O$ is $(K_f = 1.86 \ K \ kg \ mol^{-1})$ (in $^{\circ} C$)

Given that $\Delta T_f$ is the depression in freezing point of the solvent in a solution of a non-volatile solute of molality $m$,the quantity $\lim_{m \to 0} \left( \frac{\Delta T_f}{m} \right)$ is equal to:

$2.7 \ kg$ of each of water and acetic acid are mixed. The freezing point of the solution will be $-x^{\circ} C$. Consider the acetic acid does not dimerise in water,nor dissociates in water. $x = . . . . . . .$ (nearest integer)
[Given : Molar mass of water $= 18 \ g \ mol^{-1}$,acetic acid $= 60 \ g \ mol^{-1}$]
$K_f \ H_2O = 1.86 \ K \ kg \ mol^{-1}$
$K_f$ acetic acid $= 3.90 \ K \ kg \ mol^{-1}$
Freezing point: $H_2O = 273 \ K$,acetic acid $= 290 \ K$

In winter,the normal temperature in Kullu valley was found to be $-11\,^{\circ}C$. Is a $28\%$ (by mass) aqueous solution of ethylene glycol suitable for a car radiator? $K_f$ for water $= 1.86\, K\, kg\, mol^{-1}$.

Calculate the molality of a solution having freezing point depression $3.6 \ K$ and freezing point depression constant $4.8 \ K \ kg \ mol^{-1}$.