What is the freezing point depression constant of a solvent,$50 \ g$ of which contains $1 \ g$ of a non-volatile solute (molar mass $256 \ g \ mol^{-1}$) and the decrease in freezing point is $0.40 \ K$?

The addition of $0.643 \,g$ of a compound to $50 \,mL$ of benzene (density $= 0.879 \,g \,mL^{-1}$) lowers the freezing point from $5.51^{\circ}C$ to $5.03^{\circ}C$. If the freezing point constant,$K_f$ for benzene is $5.12 \,K \,kg \,mol^{-1}$,the molar mass of the compound is approximately $..... \,g \,mol^{-1}$.

$1.8 \ g$ of fructose is added to $2 \ kg$ of water. The freezing point of the solution is $(k_f = 1.86 \ K \ kg \ mol^{-1})$ (in $^\circ C$)

What is the freezing point of a $1 \ molal$ aqueous solution of a non-volatile solute (in $^{\circ} C$)? $(K_{f} = 1.86 \ K \ kg \ mol^{-1}, T_{f}^{\circ} \text{ for water } = 0^{\circ} C)$

$1.8 \ g$ of glucose (molar mass $180 \ g \ mol^{-1}$) is dissolved in $0.1 \ kg$ of water. The freezing point of the solution (in $^{\circ}C$) is ($K_f$ for water $= 1.86 \ K \ kg \ mol^{-1}$)

Find the freezing point depression of a solution having molality $0.25 \ mol \ kg^{-1}$. $(K_{f} = 4.0 \ K \ kg \ mol^{-1})$ (in $K$)