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Colligative Properties: Freezing Point And Osmosis

Find molar mass from the drop in freezing point and from osmotic pressure.

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Depression of Freezing Point

Explains freezing point depression.

The freezing point is the temperature where the vapor pressure of a liquid equals the vapor pressure of its solid phase. When a non-volatile solute is added to a solvent, the vapor pressure drops. Because of this, the solution must be cooled to a lower temperature to freeze. This decrease in freezing point is called the freezing point depression ( $\Delta T_f$ ).

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Example 1.9: Calculating FP Depression

Calculates freezing point depression for ethylene glycol.

Problem

$45\text{ g}$ of ethylene glycol ( $C_2H_6O_2$ ) is mixed with $600\text{ g}$ of water. Calculate (a) the freezing point depression and (b) the freezing point of the solution. ( $K_f$ for water is $1.86\text{ K kg mol}^{-1}$ ).

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Molar Mass from FP

Solves for molar mass of a solute in benzene.

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In Example 1.10, 1.00 g of a non-electrolyte solute is dissolved in 50 g of benzene, lowering the freezing point by 0.40 K. The freezing point depression constant ( $K_f$ ) of benzene is given as 5.12 K kg mol⁻¹.

To find the molar mass of the solute ( $M_2$ ), we use the rearranged equation 1.36: $M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$ .

Substituting the values into the numerator gives us $5.12 \times$ $\times 1000$ . For the denominator, we multiply the temperature change $\Delta T_f$ of K by the solvent mass $w_1$ of 50 g.

Solving this expression gives a final calculated molar mass of g mol⁻¹ for the unknown non-electrolyte solute.

Osmotic Pressure Diagram

Apparatus showing the pressure applied to stop osmosis.

Clean scientific diagram showing a U-shaped apparatus separated by a semipermeable membrane in the bottom middle. The left chamber contains pure solvent and the right chamber contains a solution. A piston on the right solution side pushes down with an applied pressure labeled Pi, stopping the natural flow of solvent.

Click to zoom

This setup illustrates the measurement of osmotic pressure by identifying the exact external force required to counteract natural solvent migration.

The Barrier: A semipermeable membrane separates pure solvent from a solution, allowing only small solvent molecules to pass while blocking larger solute particles.
The Process: Solvent naturally moves toward the solution side to equalize concentration, but an external pressure ( $P_i$ ) is applied via a piston to resist this flow.
The Equilibrium: When $P_i$ is exactly enough to stop the flow (net movement = 0), that value is defined as the osmotic pressure ( $\Pi$ ) of the solution.
The Threshold: This state is the "tipping point"; if $P_i$ increases further, it forces solvent back through the membrane, initiating reverse osmosis.

Osmosis and Osmotic Pressure

Defines osmosis, osmotic pressure, and isotonic solutions.

When a semipermeable membrane (SPM) separates a solvent and a solution, solvent molecules naturally flow through the membrane. Crucially, they always flow from the pure solvent (or lower concentration) into the solution (higher concentration). This spontaneous flow is called osmosis.

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Example 1.11: Molar Mass of a Protein

Calculates the massive molar mass of a protein using osmotic pressure.

Problem

$200\text{ cm}^3$ of an aqueous solution of a protein contains $1.26\text{ g}$ of the protein. The osmotic pressure of such a solution at $300\text{ K}$ is found to be $2.57 \times 10^{-3}\text{ bar}$ . Calculate the molar mass of the protein.

Why Osmotic Pressure? This method is widely preferred for macromolecules like proteins and polymers. It is measured around room temperature (preventing heat degradation) and its magnitude is large and measurable even for very dilute solutions.

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Reverse Osmosis

Explains reverse osmosis and its application in desalination.

The normal direction of osmosis can be reversed if a pressure larger than the osmotic pressure ( $\Pi$ ) is applied to the solution side.

When this happens, the pure solvent is forced out of the solution and pushed back through the semipermeable membrane. This phenomenon is called reverse osmosis.

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Depression of Freezing Point

Explains freezing point depression.

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Example 1.9: Calculating FP Depression

Calculates freezing point depression for ethylene glycol.

Problem

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Molar Mass from FP

Solves for molar mass of a solute in benzene.

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To find the molar mass of the solute ( $M_2$ ), we use the rearranged equation 1.36: $M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$ .

Substituting the values into the numerator gives us $5.12 \times$ $\times 1000$ . For the denominator, we multiply the temperature change $\Delta T_f$ of K by the solvent mass $w_1$ of 50 g.

Solving this expression gives a final calculated molar mass of g mol⁻¹ for the unknown non-electrolyte solute.

Osmotic Pressure Diagram

Apparatus showing the pressure applied to stop osmosis.

Click to zoom

This setup illustrates the measurement of osmotic pressure by identifying the exact external force required to counteract natural solvent migration.

The Barrier: A semipermeable membrane separates pure solvent from a solution, allowing only small solvent molecules to pass while blocking larger solute particles.
The Process: Solvent naturally moves toward the solution side to equalize concentration, but an external pressure ( $P_i$ ) is applied via a piston to resist this flow.
The Equilibrium: When $P_i$ is exactly enough to stop the flow (net movement = 0), that value is defined as the osmotic pressure ( $\Pi$ ) of the solution.
The Threshold: This state is the "tipping point"; if $P_i$ increases further, it forces solvent back through the membrane, initiating reverse osmosis.

Example 1.11: Molar Mass of a Protein

Calculates the massive molar mass of a protein using osmotic pressure.

Problem

Why Osmotic Pressure? This method is widely preferred for macromolecules like proteins and polymers. It is measured around room temperature (preventing heat degradation) and its magnitude is large and measurable even for very dilute solutions.

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Reverse Osmosis

Explains reverse osmosis and its application in desalination.

The normal direction of osmosis can be reversed if a pressure larger than the osmotic pressure ( $\Pi$ ) is applied to the solution side.

When this happens, the pure solvent is forced out of the solution and pushed back through the semipermeable membrane. This phenomenon is called reverse osmosis.

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