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 cryoscopic constant $(K_f)$ of a solvent when $2.5 \ g$ of a solute is dissolved in $35 \ g$ of the solvent,which lowers its freezing point by $3 \ K$. (Molar mass of the solute is $117 \ g \ mol^{-1}$)

$1 \ kg$ of $0.75 \ m$ aqueous solution of sucrose is cooled to $-4^{\circ} C$. The amount of ice (in $g$) that will be separated out is .... . (Nearest integer)
Given: $K_{f}(H_{2}O) = 1.86 \ K \ kg \ mol^{-1}$

$0.05 \ mol$ of a non-volatile solute is dissolved in $500 \ g$ of water. What is the depression in freezing point of the resultant solution (in $K$)? $(K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1})$

The $K_f$ of $1,4-dioxane$ is $4.9 \, K \, kg \, mol^{-1}$. The depression in freezing point for a $0.001 \, m$ solution in dioxane is

Calculate the molecular weight of a substance whose $7.0\%$ by mass solution in water freezes at $-0.93\,^{\circ}C$. The cryoscopic constant of water is $1.86\,^{\circ}C\,kg\,mol^{-1}$. .......... $g\,mol^{-1}$.