How does the Freezing Point Depression work?

ΔTf = i × Kf × m (molality, not molarity) Then: Kf(water) = 1.86 °C·kg/mol Then: Molality m = moles solute / kg solvent Then: van't Hoff factor i = number of particles per formula unit (1 for glucose, 2 for NaCl, 3 for CaCl₂).

Can you give an example of how to use the Freezing Point Depression?

Example: Input 1 mol NaCl in 1 kg water · i=2 · Kf=1.86 gives a result of ΔTf = 3.72°C → freezes at −3.72°C (2×1.86×1=3.72°C depression).

Is the Freezing Point Depression free to use?

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How accurate is the Freezing Point Depression?

Our Freezing Point Depression uses verified mathematical formulas and is accurate to multiple decimal places. Results are calculated in real-time using the same methods used by professionals.

Praktikal

Freezing Point Depression

❄️Freezing Point Depression (ΔTf=iKfm)

Kf (°C·kg/mol)

Water=1.86

Molality m (mol/kg)

van't Hoff factor i

NaCl=2, glucose=1

Freezing point depression is a colligative property: adding a solute to a pure solvent lowers the freezing point. ΔTf = i × Kf × m, where i is the van't Hoff factor, Kf is the cryoscopic constant, and m is molality.

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Did You Know?

Road salt (NaCl) works by freezing point depression, but only down to about −9°C. At lower temperatures, CaCl₂ (which gives i=3 and larger ΔTf) is used for de-icing.

References

›Freezing-point depression — Wikipedia

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Freezing Point Depression

❄️Freezing Point Depression (ΔTf=iKfm)

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❄️Freezing Point Depression (ΔTf=iKfm)