Grade 11

Advanced

10 min

Lesson 5 of 12

Colligative Properties of Solutions

Not Started

Investigating how the presence of a solute changes the physical properties of a solvent.

Why does a handful of salt keep a pot of water from boiling immediately, yet allow it to reach a higher temperature than pure water ever could? The answer lies not in the identity of the salt, but in the sheer number of its particles.

Learning Objectives

1.Calculate boiling point elevation and freezing point depression using molality and the van't Hoff factor.
2.Explain how the van't Hoff factor ( $i$ ) accounts for the dissociation of electrolytes in solution.
3.Analyze the relationship between vapor pressure lowering and osmotic pressure in biological systems.

The Nature of Colligative Properties

Colligative properties are physical changes in a solution that depend solely on the number of solute particles, not their chemical identity. When you add a non-volatile solute to a solvent, it interferes with the solvent's ability to transition between phases. This begins with Vapor Pressure Lowering. Solute particles occupy surface area, physically blocking solvent molecules from escaping into the gas phase. According to Raoult's Law, the vapor pressure of a solution is directly proportional to the mole fraction of the solvent:

P_{soln} = \chi_{solv} P^\circ_{solv}

Because the vapor pressure is lower, you must heat the liquid to a higher temperature to match atmospheric pressure, leading to boiling point elevation.

Quick Check

If you add 1 mole of sugar and 1 mole of urea to two separate liters of water, which will have a lower vapor pressure?

Both sugar and urea are non-electrolytes.

Answer

They will have the same vapor pressure because colligative properties depend on the number of particles, not the type of substance.

Boiling and Freezing: The Energy Shift

To quantify these changes, we use molality (

m

), which is moles of solute per kilogram of solvent. Unlike molarity, molality is temperature-independent. The formulas for boiling point elevation (

\Delta T_b

) and freezing point depression (

\Delta T_f

) are:

\Delta T_b = i K_b m

\Delta T_f = i K_f m

Here,

K_b

and

K_f

are constants specific to the solvent. The van't Hoff factor (

i

) represents the number of particles a compound dissociates into. For non-electrolytes like glucose,

i = 1

. For electrolytes like

NaCl

i = 2

because it splits into

Na^+

and

Cl^-

ions.

Calculate the freezing point depression of a solution containing $0.50$ mol of glucose ( $i=1$ ) in $1.0$ kg of water. ( $K_f$ for water is $1.86 ^\circ C/m$ ).

1. Identify variables: $i = 1$ , $m = 0.50$ mol/kg, $K_f = 1.86$ . 2. Apply formula: $\Delta T_f = (1)(1.86)(0.50) = 0.93 ^\circ C$ . 3. New freezing point: $0.00 ^\circ C - 0.93 ^\circ C = -0.93 ^\circ C$ .

Osmotic Pressure: The Biological Force

Osmosis is the movement of solvent through a semi-permeable membrane from a region of low solute concentration to high solute concentration. The pressure required to stop this flow is the Osmotic Pressure (

\Pi

). This is a critical concept in biology, as it determines how cells maintain their shape. The formula mirrors the ideal gas law:

\Pi = iMRT

where

M

is molarity,

R

is the gas constant (

0.0821 \text{ L}\\cdot\text{atm/mol}\\cdot\text{K}

), and

T

is temperature in Kelvin. High osmotic pressure in blood can lead to cellular dehydration, while low pressure can cause cells to burst.

Compare the boiling point elevation of $0.10 m$ $NaCl$ ( $i=2$ ) versus $0.10 m$ $CaCl_2$ ( $i=3$ ) in water ( $K_b = 0.512 ^\circ C/m$ ).

1. For $NaCl$ : $\Delta T_b = 2 \times 0.512 \times 0.10 = 0.1024 ^\circ C$ . 2. For $CaCl_2$ : $\Delta T_b = 3 \times 0.512 \times 0.10 = 0.1536 ^\circ C$ . 3. Conclusion: $CaCl_2$ is more effective at raising the boiling point because it produces more ions per formula unit.

Quick Check

Why is the van't Hoff factor for $MgSO_4$ often slightly less than 2 in real-world experiments?

Think about the attraction between

+2

and

-2

charges.

Answer

Ion pairing occurs, where some ions stick together in solution, effectively reducing the total number of independent particles.

A $2.00$ g sample of an unknown non-electrolyte is dissolved in enough water to make $100$ mL of solution. The osmotic pressure is $1.50$ atm at $298$ K. Find the molar mass.

1. Solve for Molarity ( $M$ ): $M = \frac{\Pi}{RT} = \frac{1.50}{(0.0821)(298)} \approx 0.0613$ mol/L. 2. Find moles in $0.100$ L: $0.0613 \times 0.100 = 0.00613$ moles. 3. Calculate Molar Mass: $\frac{2.00 \text{ g}}{0.00613 \text{ mol}} \approx 326.26$ g/mol.

Key Takeaways

1Colligative properties depend on the concentration of solute particles, not their identity.
2The van't Hoff factor ( $i$ ) is essential for electrolyte solutions; it multiplies the effect of the molality.
3Adding a solute always lowers vapor pressure, lowers the freezing point, and raises the boiling point.
4Osmotic pressure ( $\Pi = iMRT$ ) is the pressure needed to prevent osmosis across a membrane.

Quick Check

Multiple Choice

Which of the following $0.1 m$ aqueous solutions will have the lowest freezing point?

Multiple Choice

What happens to the vapor pressure of a solvent when a non-volatile solute is added?

True/False

Molality is used in colligative property formulas because it does not change with temperature.

What's Next?

Review Tomorrow

In 24 hours, try to write down the four main colligative properties and the formula for boiling point elevation from memory.

Practice Activity

Look at the ingredients of a sports drink. Identify which ingredients are electrolytes and estimate their van't Hoff factors.

Browse

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Colligative Properties of Solutions

Learning Objectives

The Nature of Colligative Properties

Boiling and Freezing: The Energy Shift

Osmotic Pressure: The Biological Force

Key Takeaways

Quick Check

What's Next?

Colligative Properties of Solutions

Learning Objectives

The Nature of Colligative Properties

Boiling and Freezing: The Energy Shift

Osmotic Pressure: The Biological Force

Key Takeaways

Quick Check

What's Next?