$45 \ g$ of ethylene glycol $(C_{2}H_{6}O_{2})$ is mixed with $600 \ g$ of water. Calculate $(a)$ the freezing point depression and $(b)$ the freezing point of the solution.

(N/A) The depression in freezing point is calculated using the formula $\Delta T_{f} = K_{f} \times m$,where $m$ is the molality of the solution.
$1.$ Calculate the moles of ethylene glycol: $\text{Moles} = \frac{45 \ g}{62 \ g \ mol^{-1}} = 0.726 \ mol$.
$2.$ Calculate the mass of water in $kg$: $\text{Mass} = \frac{600 \ g}{1000 \ g \ kg^{-1}} = 0.6 \ kg$.
$3.$ Calculate the molality $(m)$: $m = \frac{0.726 \ mol}{0.6 \ kg} = 1.21 \ mol \ kg^{-1}$.
$4.$ Calculate the freezing point depression $(\Delta T_{f})$: $\Delta T_{f} = 1.86 \ K \ kg \ mol^{-1} \times 1.21 \ mol \ kg^{-1} = 2.25 \ K$.
$5.$ Calculate the freezing point of the solution: $T_{f} = T_{f}^{\circ} - \Delta T_{f} = 273.15 \ K - 2.25 \ K = 270.90 \ K$.