Identify the correct relation between depression in freezing point and the freezing point of a pure solvent.

The freezing point of a $5\%$ (by mass) solution of cane sugar in water is $271 \, K$. If the freezing point of pure water is $273.15 \, K$,the freezing point of a $5\%$ (by mass) solution of glucose in water will be .......... $K$.

$2.7 \ kg$ 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 $= 290 \ K$

What is molal depression constant? Derive an equation.

$31 \ g$ of ethylene glycol $(C_2H_6O_2)$ is dissolved in $600 \ g$ of water. The freezing point depression of the solution is ($K_f$ for water is $1.86 \ K \ kg \ mol^{-1}$) (in $K$)

$45 \ g$ of ethylene glycol $(C_{2}H_{6}O_{2})$ is mixed with $600 \ g$ of water. Calculate $(a)$ the freezing point depression and $(b)$ the freezing point of the solution.