Calculate the molar mass of a non-volatile solute when $1 \ g$ of it is dissolved in $100 \ g$ of solvent,which decreases its freezing point by $0.2 \ K$. Given: $K_{f} = 1.2 \ K \ kg \ mol^{-1}$.

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 = $273 \ K$)

$31 \ g$ 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 $K$)

Column-$I$ (Various solutions)	Column-$II$ (Freezing point)
$a$. $0.1 \, M \ BaCl_2$ solution	$p$. $271 \, K$
$b$. $0.1 \, M \ NaCl$ solution	$q$. $270 \, K$
$c$. $0.1 \, M \ K_3[Fe(CN)_6]$ solution	$r$. $268 \, K$
$d$. $0.1 \, M \ Al_2(SO_4)_3$ solution	$s$. $269 \, K$

Given: Freezing point of $0.1 \, M$ sucrose solution $= 272 \, K$ and freezing point of water $= 273 \, K$.
Which of the following options shows the correct matches?

Given below are two statements $:$
Statement $(I) :$ Molal depression constant $K_{f}$ is given by $\frac{M_1 R T_f^2}{1000 \Delta H_{\text {fus }}}$,where symbols have their usual meaning. (Note: The provided formula in the prompt was corrected to the standard thermodynamic expression $K_f = \frac{M_1 R T_f^2}{\Delta H_{\text {fus }}}$).
Statement $(II) :$ $K_{f}$ for benzene is less than the $K_{f}$ for water.
In the light of the above statements,choose the most appropriate answer from the options given below $:$

Calculate the mass of ascorbic acid (Vitamin $C$,$C_{6}H_{8}O_{6}$) to be dissolved in $75 \ g$ of acetic acid to lower its melting point by $1.5 \ ^{\circ}C$. $K_{f} = 3.9 \ K \ kg \ mol^{-1}$.