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Colligative Properties Calculator

Colligative properties are solution properties that depend only on the number of solute particles, not on the type of solute. There are four main colligative properties: vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. Boiling point elevation (ΔTb) is calculated using ΔTb = Kb × m × i. Freezing point depression (ΔTf) uses ΔTf = Kf × m × i. Osmotic pressure (π) is calculated using π = MRT × i. Vapor pressure lowering uses ΔP = P₀ × Xt. The Van't Hoff factor (i) indicates the number of particles produced by dissociation: i = 1 for non-electrolytes (sugar), i = 2 for...

ΔTb = Kb × m × i; ΔTf = Kf × m × i; π = MRT × i; ΔP = P₀ × Xt

Variables:

  • ΔTbBoiling point elevation
    (e.g.: 0.512 °C)
  • KbBoiling point elevation constant
    (e.g.: 0.512 °C/m (water))
  • ΔTfFreezing point depression
    (e.g.: 1.86 °C)
  • KfFreezing point depression constant
    (e.g.: 1.86 °C/m (water))
  • mMolality
    (e.g.: 1 mol/kg)
  • iVan't Hoff factor
    (e.g.: 2 (NaCl))
  • πOsmotic pressure
    (e.g.: 24.5 atm)
  • MMolarity
    (e.g.: 1 M)
  • RGas constant
    (e.g.: 0.0821 L·atm/(mol·K))
  • TTemperature
    (e.g.: 298 K)

Categories:

How to Use the Calculator

  1. 1

    Select Property

    Choose the colligative property to calculate: boiling point elevation, freezing point depression, osmotic pressure, or vapor pressure.

  2. 2

    Enter Data

    Enter parameter values for the selected property.

  3. 3

    Van't Hoff Factor (i)

    Enter the Van't Hoff factor (i): 1 for non-electrolytes, 2 for NaCl, 3 for CaCl₂.

  4. 4

    Calculate

    Click calculate to get the result.

Examples

Boiling Point Elevation of Water

Problem:

A 1 molal NaCl solution in water. What is the boiling point elevation and solution boiling point? (Kb water = 0.512 °C/m)

Solution:
  1. 1.i = 2 (NaCl dissociates into Na⁺ and Cl⁻)
  2. 2.ΔTb = Kb × m × i = 0.512 × 1 × 2 = 1.024 °C
  3. 3.Solution Tb = 100 + 1.024 = 101.024 °C
Result:ΔTb = 1.024 °C, Tb = 101.024 °C

The solution boils at 101.024 °C.

Freezing Point Depression

Problem:

A 0.5 molal glucose solution in water. What is the freezing point depression? (Kf water = 1.86 °C/m)

Solution:
  1. 1.i = 1 (glucose is a non-electrolyte)
  2. 2.ΔTf = Kf × m × i = 1.86 × 0.5 × 1 = 0.93 °C
  3. 3.Solution Tf = 0 - 0.93 = -0.93 °C
Result:ΔTf = 0.93 °C, Tf = -0.93 °C

The solution freezes at -0.93 °C.

Frequently Asked Questions

Why are colligative properties important?
Colligative properties are used daily: salt to melt ice (lowers freezing point), preventing radiator freeze, osmosis in cells, seawater desalination, and determining molecular mass.
What is the Van't Hoff factor (i)?
The Van't Hoff factor is the number of particles from dissociation. Non-electrolytes (sugar, urea) have i=1. Electrolytes: NaCl→Na⁺+Cl⁻ (i=2), CaCl₂→Ca²⁺+2Cl⁻ (i=3), Al₂(SO₄)₃ (i=5).
Why is salt used to melt ice on roads?
Salt lowers the freezing point of water. Pure water freezes at 0°C, but salt solution freezes below 0°C. Sprinkling salt melts ice even when the temperature is still below freezing, making roads safer.

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References