An antifreeze solution is prepared by dissolving \(31g\) of ethylene glycol \((M=62g\,mol^{-1})\) in \(600g\) of water. Calculate the freezing point of the solution.
\[
K_f\text{ for water}=1.86K\,kg\,mol^{-1}
\]
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For depression in freezing point, use \(\Delta T_f=K_fm\), and final freezing point is \(0-\Delta T_f\) for water.
Step 1: Write the formula for depression in freezing point. \[ \Delta T_f = K_f \times m \] where $m$ = molality (mol solute per kg solvent) and $K_f$ = cryoscopic constant of solvent. Step 2: Calculate moles of ethylene glycol (solute). Molar mass of ethylene glycol = 62 g/mol (given). Mass = 31 g. \[ n_{solute} = \frac{31}{62} = 0.5 \text{ mol} \] Step 3: Calculate the molality. Mass of water (solvent) = 600 g = 0.6 kg. \[ m = \frac{n_{solute}}{\text{mass of solvent in kg}} = \frac{0.5}{0.6} = 0.833 \text{ mol kg}^{-1} \] Step 4: Calculate the depression in freezing point. Ethylene glycol is a non-electrolyte (van't Hoff factor $i = 1$). \[ \Delta T_f = 1.86 \times 0.833 = 1.55 \text{ K} \] Step 5: Find the freezing point of the solution. Freezing point of pure water = $0^\circ C$. \[ T_f = 0 - \Delta T_f = 0 - 1.55 = -1.55^\circ C \] Step 6: State the final answer. The freezing point of the antifreeze solution is $-1.55^\circ C$. This lowering of the freezing point is the reason ethylene glycol works as an antifreeze in car radiators. \[ \boxed{T_f = -1.55^\circ C} \]