Calculate the molality of a solution of a non-volatile solute having a depression in freezing point of $0.93 \ K$ and a cryoscopic constant of the solvent of $1.86 \ K \ kg \ mol^{-1}$.

$2.7 \ kg$ 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 $= 290 \ K$

Calculate the molality of the solution of a nonvolatile solute if it freezes at $-0.36 \ ^{\circ}C$. [Given: $K_{f}$ for solvent $= 1.86 \ K \ kg \ mol^{-1}$]

Statement $1$: At the freezing point,the solid substance crystallizes from the solution.
Statement $2$: Depression of freezing point is the difference between the freezing point of the solvent and the freezing point of the solution.

Identify $(i)$,$(ii)$ and $(iii)$ in the following diagram representing the depression in freezing point:

What should be the freezing point of an aqueous solution containing $17 \ g$ of $C_2H_5OH$ in $1000 \ g$ of water? (Given: $K_f$ of water = $1.86 \ K \ kg \ mol^{-1}$)