When $0.01 \ mol$ of sugar is dissolved in $100 \ g$ of a solvent,the depression in freezing point is $0.40 \ ^oC$. When $0.03 \ mol$ of glucose is dissolved in $50 \ g$ of the same solvent,the depression in freezing point will be $......... \ ^oC$.

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

The freezing point of a $1\%$ aqueous solution of calcium nitrate will be:

At $T$ $(K)$,$x \ g$ of a non-volatile solid (molar mass $78 \ g \ mol^{-1}$) when added to $0.5 \ kg$ water,lowered its freezing point by $1.0^{\circ} C$. What is $x$ (in $g$)? ($K_{f}$ of water at $T$ $(K)$ = $1.86 \ K \ kg \ mol^{-1}$)

If the freezing point of an aqueous urea solution is $271.14 \ K$ at $1 \ \text{atm}$ pressure (given $K_f$ of water = $1.86 \ K \ kg/mol$),then what is the mole fraction of urea in this solution? (Freezing point of pure water is $273 \ K$)

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 $:$