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

For water,$K_f = 1.86 \ K \ kg \ mol^{-1}$. If your car radiator contains $1.0 \ kg$ of water,how many grams of ethylene glycol $(C_2H_6O_2)$ must be added to lower the freezing point of the solution to $-2.8 \ ^\circ C$?

Which of the following formulae is used to calculate the depression in freezing point?

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 $1.86\,^{\circ}C\,kg\,mol^{-1}$. .......... $g\,mol^{-1}$.

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

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 $..... \,g \,mol^{-1}$.