Which of the following $0.10 \ m$ aqueous solutions will have the lowest freezing point?

$A$ fixed amount of glucose is dissolved in $100 \text{ g}$ water to form a solution that freezes at $-0.2^\circ\text{C}$. If the solution is cooled down to $-0.25^\circ\text{C}$, then ......... $\text{g}$ of ice would have separated.

$1.00 \ g$ of a non-electrolyte solute (molar mass $250 \ g \ mol^{-1}$) was dissolved in $51.2 \ g$ of benzene. If the freezing point depression constant,$K_f$ of benzene is $5.12 \ K \ kg \ mol^{-1},$ the freezing point of benzene will be lowered by .......... $K$.

The freezing point of a solution containing $1.25 \ g$ of a non-electrolyte solute in $20 \ g$ of water is $271.9 \ K$. What is the molar mass of the solute? (Given: $K_f$ for water = $1.86 \ K \ kg \ mol^{-1}$,Freezing point of pure water = $273 \ K$)

$2.7 \ kg$ 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 $= 290 \ K$

$1 \ kg$ of $0.75 \ m$ aqueous solution of sucrose is cooled to $-4^{\circ} C$. The amount of ice (in $g$) that will be separated out is .... . (Nearest integer)
Given: $K_{f}(H_{2}O) = 1.86 \ K \ kg \ mol^{-1}$