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 = $273 \ K$)

What should be the freezing point of an aqueous solution containing $17 \ g$ of $C_2H_5OH$ in $1000 \ g$ of water? (Given: $K_f$ of water = $1.86 \ K \ kg \ mol^{-1}$)

The freezing point of a solution containing $10 \ mL$ of non-volatile and non-electrolyte liquid $A$ in $500 \ g$ of water is $-0.413^{\circ} C$. If $K_f$ of water is $1.86 \ K \ kg \ mol^{-1}$ and the molecular weight of $A = 60 \ g \ mol^{-1}$, what is the density of the solution in $g \ mL^{-1}$? (Assume $\Delta_{\text{mix}} V = 0$)

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 freezing point of a $0.05 \ m$ aqueous solution of a non-electrolyte. (in $K$)

Calculate the molality of a solution of a non-volatile solute having a depression in freezing point of $0.93 \ K$ and a cryoscopic constant of the solvent of $1.86 \ K \ kg \ mol^{-1}$.