The freezing point of a $0.05 \ molal$ solution of a non-electrolyte in water is $(K_f = 1.86 \ K \ kg \ mol^{-1})$:

$A$ solution of sucrose (molar mass $= 342 \, g \, mol^{-1}$) has been prepared by dissolving $68.5 \, g$ of sucrose in $1000 \, g$ of water. The freezing point of the solution obtained will be ......... $^oC$. ($K_f$ for water $= 1.86 \, K \, kg \, mol^{-1}$)

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 $:$

For $1000 \ g$ of $1,4$-dioxane,the cryoscopic constant $K_f = 4.9 \ K \ kg \ mol^{-1}$. What will be the depression in freezing point for a $0.001 \ m$ solution prepared in dioxane?

$2 \ g$ of a non-electrolyte solute (molar mass is $500 \ g \ mol^{-1}$) was dissolved in $57.3 \ g$ of xylene. If the freezing point depression constant $K_f$ of xylene is $4.3 \ K \ kg \ mol^{-1}$,then the depression in freezing point of xylene is.......... (in $K$)

$50 \ g$ of antifreeze (ethylene glycol) is added to $200 \ g$ of water. What amount of ice will separate out at $-9.3 \ ^oC$? $(K_f = 1.86 \ K \ kg \ mol^{-1})$