Freezing point depression calculator

How do you calculate freezing point depression?

ΔTf = Kf × m, where Kf is the solvent’s freezing point depression constant and m the molality of the dissolved particles. Boiling point elevation works the same way: ΔTb = Kb × m (OpenStax Chemistry 2e).

Example: salt water

OpenStax works out 4.2 g of NaCl in 125 g of water: 0.072 mol of NaCl gives 0.14 mol of ions, 1.2 m. ΔTf = 1.86 × 1.2 ≈ 2.1 °C, so the solution freezes at about −2.1 °C.

What is the van ’t Hoff factor i?

The number of moles of particles in solution per mole of solute dissolved. It is 1 for non-electrolytes such as sugar and 2 in theory for NaCl; the measured value for 0.050 m NaCl is 1.9, because some ions stay paired.

What are Kf and Kb for water?

Kf = 1.86 °C·kg/mol and Kb = 0.512 °C·kg/mol. The calculator also has glacial acetic acid, benzene, chloroform and nitrobenzene from the same OpenStax table.

Example: boiling point

OpenStax dissolves 92.1 g of iodine (I₂) in 800.0 g of chloroform: 0.454 m, ΔTb = 3.63 × 0.454 = 1.65 °C, so the solution boils at 62.91 °C.

Why is the real effect of salt smaller than the calculation?

The formula assumes the ideal van 't Hoff factor, i = 2 for NaCl and i = 3 for CaCl₂, which are the theoretical maximums. In real solutions ions pair up, so the measured factor is lower, and at high concentrations the dilute-solution formula no longer holds. For a 1 m NaCl solution the ideal answer is a 3.72 °C drop (6.7 °F), and for CaCl₂ 5.58 °C; real solutions show somewhat less.

Does freezing point depression depend on what the solute is?

Only through the number of dissolved particles. That is what makes it a colligative property: 1 mol of any non-dissociating solute in 1 kg of water lowers the freezing point by the same 1.86 °C, whether it is sugar, glycerol or urea. A solute that splits into ions counts each ion.

The formulas and their limits

ΔTf = i × Kf × m lowers the freezing point and ΔTb = i × Kb × m raises the boiling point, where m is the molality (moles of solute per kilogram of solvent), Kf and Kb are properties of the solvent and i is the van 't Hoff factor. Molality, not molarity, is used because it does not change with temperature. The equations apply to dilute solutions of a non-volatile solute, and they assume the solid that freezes out is pure solvent.

Worked example: sugar versus salt in 1 kg of water

Dissolving 0.50 mol of sucrose (about 171 g) in 1.00 kg of water gives m = 0.50 and i = 1, so ΔTf = 1.86 × 0.50 = 0.93 °C and the solution freezes at -0.93 °C. The boiling point rises by 0.512 × 0.50 = 0.256 °C to 100.256 °C. The same 0.50 mol of NaCl gives about 1.0 mol of ions, so the ideal ΔTf is 1.86 °C, twice as large. In benzene, whose Kf is 5.12, 0.50 m of a non-electrolyte lowers the freezing point by 2.56 °C to 2.94 °C.

Common mistakes

Using the mass of the solution instead of the mass of the solvent in the molality, or the solute's mass in grams without converting it to moles, gives the wrong m. Forgetting the van 't Hoff factor for salts is the next: it is 1 for sugar but 2 for NaCl in ideal solutions. Using molarity (mol per liter of solution) in place of molality is a smaller error in dilute water solutions but a real one in concentrated ones. Finally, a temperature change from the formula is a difference: the freezing point is the pure solvent's freezing point minus ΔTf, not ΔTf itself.

Sources: OpenStax Chemistry 2e, section 11.4 (colligative properties of solutions, Table 11.2 with Kf and Kb). Related: molality calculator, osmotic pressure calculator, molarity calculator and molar mass calculator.