Case 1
- Van 't Hoff factor i
- 2
- Cryoscopic constant Kf
- 1.86K·kg/mol
- Molality b
- 1mol/kg
3.720K
Open with these valuesFreezing Point Depression Calculator
3.720K
Result: 3.720 KDissolved particles lower the freezing point: ΔTf = i × Kf × b. One molal table salt in water gives 2 × 1.86 × 1 = 3.72 K, so it freezes at about −3.72 °C. The result is a difference, so kelvin and degrees Celsius are the same number here.
The numbers at a glance
Held fixed: Van 't Hoff factor i 2.00, Cryoscopic constant Kf 1.860 K·kg/mol.
| Molality b (mol/kg) | Result (K) |
|---|---|
| 0.000 | 0.000 |
| 0.250 | 0.930 |
| 0.500 | 1.860 |
| 0.750 | 2.790 |
| 1.000Your value | 3.720 |
| 1.250 | 4.650 |
| 1.500 | 5.580 |
| 1.750 | 6.510 |
| 2.000 | 7.440 |
Worked examples
Case 2
- Van 't Hoff factor i
- 1
- Cryoscopic constant Kf
- 1.86K·kg/mol
- Molality b
- 1mol/kg
1.860K
Open with these valuesCase 3
- Van 't Hoff factor i
- 3
- Cryoscopic constant Kf
- 1.86K·kg/mol
- Molality b
- 0.5mol/kg
2.790K
Open with these valuesHow it's calculated
ΔTf = i × Kf × b
- StepEnter how many particles one formula unit splits into: 1 sugar, 2 NaCl, 3 CaCl₂.
- StepEnter the cryoscopic constant of the solvent; water is 1.86 K·kg/mol.
- StepEnter the molality — moles of solute per kilogram of solvent.
- ResultSubtract the depression from the solvent's own freezing point.
Reference table
| i, Kf, b | What is dissolved | Depression in K |
|---|---|---|
| 2, 1.86, 0 | nothing — pure water | 0.000 |
| 1, 1.86, 1 | 1 molal sugar in water | 1.860 |
| 1, 5.12, 0.2 | 0.2 molal in benzene, Kf 5.12 | 1.024 |
| 3, 1.86, 0.5 | 0.5 molal calcium chloride | 2.790 |
| 2, 1.86, 1 | 1 molal table salt in water | 3.720 |
| 2, 1.86, 2.5 | 2.5 molal table salt | 9.300 |
Questions
How do I calculate freezing point depression?
Multiply the van 't Hoff factor by the cryoscopic constant and the molality: ΔTf = i × Kf × b. One molal NaCl in water is 2 × 1.86 × 1 = 3.72 K, so the solution freezes 3.72 °C below pure water.
What is the cryoscopic constant Kf?
It is a property of the solvent that says how strongly each unit of molality pulls the freezing point down. Water is about 1.86 K·kg/mol, benzene about 5.12 and camphor around 40, so always use the constant of the solvent you actually have.
What is the van 't Hoff factor i?
It is how many particles one formula unit breaks into when it dissolves: 1 for non-electrolytes such as sugar, 2 for NaCl, 3 for CaCl₂. More particles mean a bigger drop, which is why the calculator asks rather than assuming a substance.
Why does salt melt ice on roads?
Salt dissolves into Na⁺ and Cl⁻, and those extra particles lower the freezing point of the water. With enough salt, ice keeps melting several degrees below 0 °C.
Why molality and not molarity?
Molality counts moles per kilogram of solvent, and a mass does not change when the sample cools. Molarity is per litre of solution, and that volume shrinks on the way to freezing.
Sources and last check
Information, not professional advice.