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Chemistry: Matter, Mixtures, and Separation - Solutions, Suspensions, Colloids, Emulsions, and the Tyndall Effect

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Matter and Mixtures: Matter can be classified into pure substances and mixtures. A mixture is a physical combination of two or more substances where each substance retains its chemical identity.

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Solutions: A solution is a homogeneous mixture consisting of a solute dissolved in a solvent. The particle size is less than 1 nm1 \text{ nm} (10−9 m10^{-9} \text{ m}). They are stable and do not show the Tyndall effect.

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Suspensions: These are heterogeneous mixtures containing large particles (greater than 1000 nm1000 \text{ nm}) that settle down when left undisturbed (sedimentation). They can be separated by filtration.

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Colloids: These are mixtures where particle sizes are between 1 nm1 \text{ nm} and 1000 nm1000 \text{ nm}. They appear homogeneous to the naked eye but are heterogeneous under a microscope. Particles do not settle.

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Tyndall Effect: This is the phenomenon of scattering of a beam of light by particles in a colloid or a very fine suspension. This makes the path of light visible (e.g.e.g., sunlight passing through a canopy of trees).

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Emulsions: A type of colloid where both the dispersed phase and the dispersion medium are liquids (e.g.e.g., milk, mayonnaise). An emulsifying agent is often required to stabilize the mixture.

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Solubility: The maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature, usually expressed in gg per 100 g100 \text{ g} of solvent.

📐Formulae

Concentration (Mass/Mass %)=(Mass of SoluteMass of Solution)×100\text{Concentration (Mass/Mass \%)} = \left( \frac{\text{Mass of Solute}}{\text{Mass of Solution}} \right) \times 100

Mass of Solution=Mass of Solute+Mass of Solvent\text{Mass of Solution} = \text{Mass of Solute} + \text{Mass of Solvent}

Concentration (Mass/Volume %)=(Mass of SoluteVolume of Solution)×100\text{Concentration (Mass/Volume \%)} = \left( \frac{\text{Mass of Solute}}{\text{Volume of Solution}} \right) \times 100

💡Examples

Problem 1:

A solution contains 40 g40 \text{ g} of common salt in 320 g320 \text{ g} of water. Calculate the concentration in terms of mass by mass percentage of the solution.

Solution:

Mass of solute (salt) = 40 g40 \text{ g}. Mass of solvent (water) = 320 g320 \text{ g}. Total mass of solution = 40 g+320 g=360 g40 \text{ g} + 320 \text{ g} = 360 \text{ g}. Concentration =40360×100=11.11%= \frac{40}{360} \times 100 = 11.11\%.

Explanation:

To find the mass percentage, we divide the mass of the solute by the total mass of the solution (solute + solvent) and multiply by 100.

Problem 2:

Explain why the path of light is visible when a laser is shone through a glass of milk but not through a glass of salt water.

Solution:

Milk is a colloid, whereas salt water is a true solution.

Explanation:

In milk, the particles are large enough (1 nm1 \text{ nm} to 1000 nm1000 \text{ nm}) to scatter light, which is known as the Tyndall Effect. In salt water, the solute particles are less than 1 nm1 \text{ nm} and are too small to scatter light beams.

Problem 3:

Calculate the mass of potassium nitrate needed to create a saturated solution in 50 g50 \text{ g} of water at 313 K313 \text{ K}, given its solubility is 62 g62 \text{ g} per 100 g100 \text{ g} of water at that temperature.

Solution:

Solubility in 100 g100 \text{ g} water = 62 g62 \text{ g}. For 50 g50 \text{ g} water, mass =62100×50=31 g= \frac{62}{100} \times 50 = 31 \text{ g}.

Explanation:

Since 50 g50 \text{ g} is half of 100 g100 \text{ g}, the amount of solute required is also halved to maintain saturation at the same temperature.