Colligative Properties Calculator

Results

Real electrolytes show a slightly smaller shift than the ideal i predicts, because ions attract one another: NaCl measures about i = 1.9 at 0.050 m, not 2. Use a "measured" value for a closer estimate.

How to Use the Colligative Properties Calculator

1. 1. Choose the mode. "Boiling or freezing point" gives the new temperature; "Molar mass from ΔT" works backwards from a measured freezing or boiling point; "Osmotic pressure" applies Π = iMRT.
2. 2. Pick the solvent. Water, benzene, chloroform, acetic acid and nitrobenzene have built-in constants; choose "another solvent" and type its Kb or Kf and pure-solvent point if yours is not listed.
3. 3. Enter the concentration. Give the molality directly, or the masses of solute and solvent plus the solute's molar mass. For osmotic pressure use molarity, because it is per litre of solution.
4. 4. Set the van 't Hoff factor. Use 1 for a molecular solute such as sugar, 2 for NaCl, 3 for CaCl₂, or choose a "measured" value to allow for ion pairing.

The Equations

ΔTb = i × Kb × m

ΔTf = i × Kf × m

Π = i × M × R × T

A colligative property depends on how many solute particles are dissolved, not on what they are. m is the molality (mol solute per kg solvent), M the molarity (mol per litre of solution), and i the van 't Hoff factor — the number of particles each formula unit gives. K is a property of the solvent alone. A solute raises the boiling point and lowers the freezing point; osmotic pressure is the pressure needed to stop solvent flowing across a semipermeable membrane into the solution. For Π, T must be in kelvin, and R = 8.314 462 618 J/(mol·K), or 0.082 057 366 L·atm/(mol·K), is exact.

Solvent Constants

From OpenStax Chemistry 2e, Table 11.2 (normal boiling and freezing points at 1 atm; constants in °C·kg/mol):

The van 't Hoff Factor

Sugar stays whole in water, so i = 1. NaCl splits into Na⁺ and Cl⁻, so the ideal i is 2, and CaCl₂ gives 3. Real electrolytes fall short of this because oppositely charged ions pair up, and the shortfall grows with concentration. OpenStax's Table 11.3 gives the measured values for 0.050 m aqueous solutions:

Worked Examples

Salt water freezes lower (the default inputs). A 1.00 m solution of NaCl in water, taking i = 2: ΔTf = 2 × 1.86 × 1.00 = 3.72 °C, so it freezes at −3.72 °C. A real 1 m NaCl solution gives a measured shift of about 3.4 °C, because i is nearer 1.8 than 2.

Boiling point elevation. A 0.330 m solution of a nonvolatile molecular solute in benzene: ΔTb = 1 × 2.53 × 0.330 = 0.835 °C, so it boils at 80.1 + 0.835 = 80.9 °C. For iodine in chloroform, 92.1 g of I₂ (253.8 g/mol) in 800.0 g of CHCl₃ is 0.454 m, ΔTb = 3.63 × 0.454 = 1.65 °C, and the solution boils at 62.9 °C.

Molar mass from freezing-point depression. 4.00 g of an unknown molecular solute dissolved in 55.0 g of benzene freezes at 2.32 °C instead of 5.5 °C. ΔTf = 3.18 °C, m = 3.18 ÷ 5.12 = 0.621 mol/kg, n = 0.621 × 0.0550 kg = 0.0342 mol, and M = 4.00 g ÷ 0.0342 mol = 117 g/mol.

Osmotic pressure. A 0.30 M glucose solution at 37 °C (310.15 K, i = 1): Π = 1 × 300 mol/m³ × 8.314 462 618 × 310.15 = 7.7 × 10⁵ Pa = 7.6 atm, close to the roughly 7.7 atm osmotic pressure of blood. For a protein, 10.0 g in 0.500 L giving 5.9 torr at 22 °C is 3.2 × 10⁻⁴ M, so the molar mass is about 6.2 × 10⁴ g/mol — osmotic pressure suits large molecules because the pressure is easy to measure even when the molarity is tiny.

Common Mistakes

• Using molarity instead of molality. The freezing and boiling formulas use moles per kilogram of solvent. The osmotic-pressure formula uses moles per litre of solution.
• Forgetting i. 1.00 m NaCl lowers the freezing point about twice as much as 1.00 m sugar.
• Using Celsius in Π. Temperature in Π = iMRT is in kelvin; the calculator converts and rejects values at or below absolute zero.
• Treating a volatile solute as an elevation. The boiling-point formula assumes the solute is nonvolatile. If it evaporates too, the boiling point may not rise.
• Using the formulas at high concentration. They are limiting laws for dilute solutions. Above about 1 m, or for strong electrolytes, expect departures from the ideal result.

Frequently Asked Questions

What are colligative properties?

Properties of a solution that depend on the number of dissolved particles rather than their identity: vapor-pressure lowering, boiling-point elevation, freezing-point depression and osmotic pressure. Each is proportional to the solute's concentration.

How do you calculate freezing point depression?

ΔTf = i × Kf × m. Multiply the van 't Hoff factor by the solvent's freezing-point constant and the molality, then subtract the result from the pure solvent's freezing point. For 1.00 m sugar in water, ΔTf = 1.86 °C, so it freezes at −1.86 °C.

What is the van 't Hoff factor?

The number of particles a formula unit produces in solution: 1 for glucose, 2 for NaCl, 3 for CaCl₂. Measured values are a little lower than these ideal ones because ions pair up, especially in concentrated solutions.

Why does salt melt ice on roads?

Dissolving salt in the thin film of water on ice lowers its freezing point, so ice melts at temperatures below 0 °C. The effect is limited: a saturated NaCl solution cannot freeze above about −21 °C, so salt stops working in severe cold.

How do I find molar mass from freezing-point depression?

Work out ΔT from the pure and observed freezing points, then m = ΔT ÷ (i × Kf). Multiply m by the kilograms of solvent to get moles of solute, and divide the solute's mass by those moles.

What is the difference between molarity and molality?

Molarity is moles per litre of solution and changes with temperature because the volume does. Molality is moles per kilogram of solvent and does not. Boiling and freezing calculations use molality; osmotic pressure uses molarity.