Salt on an icy road melts the ice; antifreeze keeps a car's coolant liquid in winter and from boiling over in summer. Both follow from Vapor Pressure Lowering: dissolved particles lower the solvent's vapor pressure, and that shifts its freezing and boiling points. Count particles with Concentrations of Ions.
Key idea
$$\Delta T_f = i K_f m, \qquad \Delta T_b = i K_b m$$ $m$ is the molality, $i$ is the van 't Hoff factor (particles per formula unit: 1 for sugar, 2 for NaCl, 3 for $\mathrm{CaCl_2}$), and for water $K_f = 1.86$ and $K_b = 0.512\ ^\circ\text{C·kg/mol}$. The freezing point goes down by $\Delta T_f$; the boiling point goes up by $\Delta T_b$.
Why it works
A liquid boils when its vapor pressure reaches the outside pressure. A solution's vapor pressure is lower at every temperature, so it must be heated higher to reach 1 atm: the boiling point rises. At the freezing point, the liquid and the solid must have equal vapor pressures (equal tendency to escape); the solution's lowered curve meets the ice's curve at a lower temperature, so it freezes colder. Like the vapor-pressure lowering, both shifts depend only on the number of dissolved particles, which is why molality times $i$ appears.
Worked example: Sugar in water
Find the freezing point of a $0.500\ \text{mol/kg}$ glucose solution in water.
- Glucose is a nonelectrolyte: $i = 1$.
- $\Delta T_f = (1)(1.86)(0.500) = 0.930^\circ\text{C}$.
- Freezing point: $0.00 - 0.930 = -0.930^\circ\text{C}$.
Answer: $-0.930^\circ\text{C}$
Worked example: A salt with three ions
Find the freezing and boiling points of $0.200\ \text{mol/kg}\ \mathrm{CaCl_2}$ in water, treating it as fully split into ions.
- $\mathrm{CaCl_2} \to \mathrm{Ca^{2+}} + 2\,\mathrm{Cl^-}$: $i = 3$.
- $\Delta T_f = (3)(1.86)(0.200) = 1.12^\circ\text{C}$, so it freezes at $-1.12^\circ\text{C}$.
- $\Delta T_b = (3)(0.512)(0.200) = 0.307^\circ\text{C}$, so it boils at $100.31^\circ\text{C}$.
Answer: $-1.12^\circ\text{C}$ and $100.31^\circ\text{C}$.
Worked example: From grams
$25.0\ \text{g}$ of NaCl ($58.44\ \text{g/mol}$) is dissolved in $500.0\ \text{g}$ of water. Estimate the freezing point.
- Moles: $25.0/58.44 = 0.4278$; molality $0.4278/0.5000 = 0.8556\ \text{mol/kg}$.
- $\Delta T_f = (2)(1.86)(0.8556) = 3.18^\circ\text{C}$.
- Freezing point: $-3.18^\circ\text{C}$.
Answer: About $-3.18^\circ\text{C}$.
Common mistake
Three slips: forgetting the van 't Hoff factor (NaCl makes two particles), using molarity instead of molality, and adding $\Delta T_f$ to the freezing point. Dissolved particles always lower the freezing point. Check: a solution freezes below and boils above the pure solvent. (Real salts give slightly smaller shifts than $i$ predicts, because some ions pair up in solution; $i$ is an ideal upper limit.)