$CaCl_2$ is preferred over $NaCl$ for clearing ice on roads,particularly in very cold countries. This is because:

The mass of ascorbic acid $(C_6H_8O_6)$ to be dissolved in $100 \ g$ of acetic acid to lower its freezing point by $1.5^{\circ}C$ in $g$ is: (Given: $K_f$ for acetic acid $= 3.9 \ K \ kg \ mol^{-1}$)

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 $:$

$A$ solution of a nonvolatile solute is obtained by dissolving $15 \ g$ in $200 \ mL$ of water,which has a depression in freezing point of $0.75 \ K$. Calculate the molar mass of the solute if the cryoscopic constant of water is $1.86 \ K \ kg \ mol^{-1}$.

For water,$K_f = 1.86 \ K \ kg \ mol^{-1}$. If your car radiator contains $1.0 \ kg$ of water,how many grams of ethylene glycol $(C_2H_6O_2)$ must be added to lower the freezing point of the solution to $-2.8 \ ^\circ C$?

Elements $X$ and $Y$ form two non-volatile compounds ($XY$ and $XY_3$). When $10 \ g$ of $XY$ is dissolved in $50 \ g$ of ethanol,the depression in freezing point $(\Delta T_{f})$ is $5.333 \ K$. When $10 \ g$ of $XY_3$ is dissolved in $50 \ g$ of ethanol,the $\Delta T_{f}$ is $2.2857 \ K$. What are the atomic weights of $X$ and $Y$ respectively? $(K_{f} = 2 \ K \ kg \ mol^{-1})$