What is the depression of freezing point,when mole fraction of non-electrolyte solute in aqueous solution is $0.01$ (in $K$)? ($K_f$ of $H_2O = 1.86 \ K \ kg \ mol^{-1}$)

Calculate the mass of ascorbic acid ($Vitamin \ C$,$C_6H_8O_6$) to be dissolved in $75 \ g$ of acetic acid to lower its melting point by $1.5^{\circ} C$. $\left[K_f = 3 \ K \ kg \ mol^{-1}\right]$ (in $g$)

$A$ $5\%$ solution (by mass) of cane sugar in water has a freezing point of $271 \ K$ and the freezing point of pure water is $273.15 \ K$. The freezing point of a $5\%$ solution (by mass) of glucose in water is ............. $K$.

Calculate the molar mass of the solute when $1.5 \ g$ of a non-volatile solute is dissolved in $100 \ mL$ of a solvent having a density of $0.8 \ g \ mL^{-1}$,which lowers its freezing point by $0.75 \ K$. (Freezing point depression constant for the solvent is $5 \ K \ kg \ mol^{-1}$).

Arrange the following solutions in order of decreasing freezing points:
$(a) \ 0.075 \ M \ CuSO_4$ $(b) \ 0.060 \ M \ (NH_4)_2SO_4$
$(c) \ 0.14 \ M \ urea$ $(d) \ 0.04 \ M \ MgCl_2$

How much glucose $(molecular \ weight = 180 \ g/mol)$ should be added to $200 \ g \ H_2O$ so that when the solution is cooled to $-0.5^{\circ}C$,$14 \ g$ of ice separates out of the solution: [$K_f = 1.86 \ K \ kg/mol$ and melting point of $H_2O = 0^{\circ}C$] (in $g$)