7.3 Colligative Properties & Raoult's Law

7.3 Colligative Properties & Raoult's Law

Quick Summary: Colligative properties depend strictly on the number of dissolved solute particles relative to solvent molecules, not on solute identity, size, or charge. The four classic colligative properties are vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. Solute particles increase liquid entropy, stabilizing the liquid phase and depressing vapor pressure via Raoult's law ($P_{\text{soln}} = X_{\text{solvent}} P^\circ$). Colligative measurements—especially osmotic pressure—provide an exceptionally sensitive technique for determining macromolecular molar masses.

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1. Fundamentals of Colligative Properties

The term colligative comes from the Latin colligatus ("bound together"). Colligative properties reflect the collective concentration of solute particles in a solution. Dissolving $1.0\text{ mol}$ of glucose, sucrose, or urea in $1.0\text{ kg}$ of water lowers the freezing point by identical amounts ($1.86^\circ\text{C}$), regardless of molecular weight or geometry.

Thermodynamic Foundation: Entropy Stabilization

In pure liquid solvent, molecules possess lower entropy than in the gas phase. Introducing a solute generates a disordered mixture with higher entropy than pure liquid. This entropic stabilization lowers the chemical potential of the liquid solvent. Because the solution is already thermodynamically stabilized, its escaping tendency into the vapor phase is reduced (lowering vapor pressure). Similarly, freezing requires organizing solvent molecules into an ordered solid lattice; the higher entropy of the solution demands cooling to lower temperatures to achieve crystallization.

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2. Vapor Pressure Lowering & Raoult's Law

Nonvolatile Solutes

When a nonvolatile solute dissolves in a volatile solvent, solute particles occupy surface area and entropy lowers the solvent's escaping tendency. Raoult's Law states that solution vapor pressure ($P_{\text{soln}}$) equals the mole fraction of solvent ($X_{\text{solvent}}$) multiplied by the vapor pressure of pure solvent ($P^\circ_{\text{solvent}}$): $P_{\text{soln}} = X_{\text{solvent}} P^\circ_{\text{solvent}}$ Since $X_{\text{solvent}} = 1 - X_{\text{solute}}$, vapor pressure lowering ($\Delta P$) is: $\Delta P = P^\circ_{\text{solvent}} - P_{\text{soln}} = X_{\text{solute}} P^\circ_{\text{solvent}}$

Ideal vs Non-Ideal Solutions

• Ideal Solutions: Obey Raoult's law across all compositions. Solute-solute, solvent-solvent, and solute-solvent attractions are identical ($\Delta H_{\text{mix}} = 0$, $\Delta V_{\text{mix}} = 0$, e.g., benzene and toluene).
• Negative Deviations: Solute-solvent attractions exceed attractions between like molecules ($\Delta H_{\text{mix}} < 0$). Molecules are held more tightly in the liquid, depressing vapor pressure below Raoult's law predictions (e.g., acetone and chloroform forming hydrogen bonds).
• Positive Deviations: Solute-solvent attractions are weaker than cohesive forces in pure components ($\Delta H_{\text{mix}} > 0$). Molecules escape more readily, producing vapor pressures exceeding Raoult's law predictions (e.g., ethanol and hexane).

Volatile Mixtures & Fractional Distillation

When both components $A$ and $B$ are volatile: $P_{\text{total}} = P_A + P_B = X_A P^\circ_A + X_B P^\circ_B$ The vapor mole fraction of $A$ ($y_A$) is $y_A = P_A / P_{\text{total}}$. If $A$ is more volatile ($P^\circ_A > P^\circ_B$), the vapor is enriched in $A$ ($y_A > X_A$). Condensing this vapor and revaporizing it across multiple theoretical plates in a fractionating column enables separation of liquids with similar boiling points (fractional distillation).

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3. Boiling Point Elevation & Freezing Point Depression

Boiling Point Elevation

Because nonvolatile solutes lower vapor pressure, the solution must be heated to a higher temperature before vapor pressure equals atmospheric pressure ($1.00\text{ atm}$): $\Delta T_b = T_b - T_b^\circ = i K_b m$ where $K_b$ is the molal boiling point elevation constant (for water, $K_b = 0.512^\circ\text{C/m}$). Solution boiling point is $100.0^\circ\text{C} + \Delta T_b$.

Freezing Point Depression

Solute particles disrupt crystalline lattice formation, depressing the freezing point: $\Delta T_f = T_f^\circ - T_f = i K_f m$ where $K_f$ is the molal freezing point depression constant (for water, $K_f = 1.86^\circ\text{C/m}$). Solution freezing point is $0.0^\circ\text{C} - \Delta T_f$. Practical examples include spreading $\text{NaCl}$ or $\text{CaCl}_2$ on winter roads and using ethylene glycol in automotive radiators.

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4. Osmotic Pressure & the van 't Hoff Equation

Osmosis is the spontaneous flow of solvent across a semipermeable membrane from lower solute concentration (higher solvent activity) into higher solute concentration. Osmotic pressure ($\Pi$) is the external pressure required to halt net solvent transfer: $\Pi = i M R T$ where $M$ is molarity, $R = 0.08206\text{ L}\cdot\text{atm/(mol}\cdot\text{K)}$, and $T$ is temperature in Kelvin.

Biological & Industrial Applications

• Reverse Osmosis: Applying external pressure exceeding osmotic pressure ($P > \Pi$) drives solvent backward across the membrane, desalinating seawater.
• Tonicity: In isotonic solutions, cell volume is stable. In hypotonic solutions, water enters the cell, causing swelling and burst (hemolysis). In hypertonic solutions, water exits the cell, causing cellular shriveling (crenation).

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5. Colligative Formulas & Water Constants

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6. Worked Example: Protein Molar Mass via Osmotic Pressure

Osmotic pressure is ideal for macromolecules (proteins, polymers) because tiny molar concentrations yield negligible $\Delta T_f$ ($<0.001^\circ\text{C}$) but produce substantial, easily measurable osmotic heads ($>10\text{ cm } \text{H}_2\text{O}$).

Problem: A $3.50\text{ g}$ sample of an unknown enzyme (nonelectrolyte, $i = 1$) is dissolved in water to make $250.0\text{ mL}$ of solution at 25.0 °C ($298.15\text{ K}$). The solution exhibits an osmotic pressure of $0.0125\text{ atm}$. Find the molar mass.

Step 1: Calculate molarity $M = \frac{\Pi}{R T} = \frac{0.0125\text{ atm}}{(0.08206\text{ L}\cdot\text{atm/(mol}\cdot\text{K)}) \times 298.15\text{ K}} = 5.11 \times 10^{-4}\text{ mol/L}$

Step 2: Calculate moles in $250.0\text{ mL}$ ($0.2500\text{ L}$) $n = (5.11 \times 10^{-4}\text{ mol/L}) \times 0.2500\text{ L} = 1.277 \times 10^{-4}\text{ mol}$

Step 3: Calculate molar mass $M_w = \frac{3.50\text{ g}}{1.277 \times 10^{-4}\text{ mol}} = 27,400\text{ g/mol}$