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Nature of Matter: Elements, Compounds, and Mixtures - Types of mixtures

Grade 8CBSE

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

🔑Concepts

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A mixture consists of two or more substances (elements or compounds) mixed together without any chemical bond. The components of a mixture retain their original properties.

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Homogeneous Mixtures: These have a uniform composition throughout. The particles are smaller than 10−910^{-9} m (11 nm) in diameter. Examples include sugar in water, air, and alloys like brass.

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Heterogeneous Mixtures: These have a non-uniform composition where the components can often be seen with the naked eye or under a microscope. Examples include sand and salt, oil and water.

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Suspensions: A heterogeneous mixture in which the solute particles do not dissolve but remain suspended throughout the bulk of the medium. Particle size is greater than 10001000 nm. They show the Tyndall effect when particles are suspended, but settle down over time.

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Colloids: A heterogeneous mixture that appears homogeneous because the particle size (11 nm to 10001000 nm) is too small to be seen by the naked eye. They are stable and exhibit the Tyndall effect (scattering of a beam of light).

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Tyndall Effect: The phenomenon of scattering of a beam of light by colloidal particles or particles in a fine suspension, making the path of light visible.

📐Formulae

Mass by mass percentage of a solution=Mass of soluteMass of solution×100\text{Mass by mass percentage of a solution} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100

Mass by volume percentage of a solution=Mass of soluteVolume of solution×100\text{Mass by volume percentage of a solution} = \frac{\text{Mass of solute}}{\text{Volume of solution}} \times 100

Mass of solution=Mass of solute+Mass of solvent\text{Mass of solution} = \text{Mass of solute} + \text{Mass of solvent}

💡Examples

Problem 1:

A solution contains 5050 g of common salt in 450450 g of water. Calculate the concentration in terms of mass by mass percentage of the solution.

Solution:

Mass of solute (salt)=50 g\text{Mass of solute (salt)} = 50\text{ g} Mass of solvent (water)=450 g\text{Mass of solvent (water)} = 450\text{ g} Mass of solution=\text{Mass of solution} = \begin{array}{r} 50 \ + 450 \ \hline 500 \end{array} g \text{ g} Concentration=50500×100=10%\text{Concentration} = \frac{50}{500} \times 100 = 10\%

Explanation:

To find the mass percentage, we first calculate the total mass of the solution by adding the solute and solvent. Then, we divide the mass of the solute by the total mass and multiply by 100100.

Problem 2:

Identify the type of mixture for the following: (i) Lemonade, (ii) Chalk powder in water, (iii) Milk.

Solution:

(i) Lemonade is a Homogeneous Mixture (Solution). (ii) Chalk powder in water is a Suspension (Heterogeneous). (iii) Milk is a Colloid (Heterogeneous).

Explanation:

Lemonade has a uniform composition. Chalk powder does not dissolve and settles at the bottom, making it a suspension. Milk appears uniform but contains tiny droplets of fat and protein dispersed in water, making it a colloid.