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Is Matter Around Us Pure - Suspensions and Colloids

Grade 9CBSE

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

🔑Concepts

A suspension is a heterogeneous mixture in which the solute particles do not dissolve but remain suspended throughout the bulk of the medium. The particles of a suspension are visible to the naked eye and have a diameter greater than 107 m10^{-7} \text{ m} (or 100 nm100 \text{ nm}).

Suspensions are unstable; if left undisturbed, the particles settle down. They can be separated from the mixture by the process of filtration. They scatter a beam of light passing through them (Tyndall Effect) until the particles settle down.

A colloid is a heterogeneous mixture where the particle size is intermediate between a true solution and a suspension, typically between 1 nm1 \text{ nm} and 100 nm100 \text{ nm} (109 m10^{-9} \text{ m} to 107 m10^{-7} \text{ m}).

Colloidal particles are too small to be seen individually by the naked eye but are big enough to scatter a beam of light. This scattering of light is known as the Tyndall Effect.

Colloids are quite stable; their particles do not settle down when left undisturbed. They cannot be separated by filtration but can be separated by a special technique known as centrifugation.

The components of a colloidal solution are the dispersed phase (the solute-like component) and the dispersion medium (the solvent-like medium).

Common examples of colloids include Aerosols (Fog, Clouds), Foam (Shaving cream), Emulsions (Milk, Face cream), and Sols (Milk of magnesia, Mud).

📐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:

To make a saturated solution, 36 g36 \text{ g} of sodium chloride (NaClNaCl) is dissolved in 100 g100 \text{ g} of water at 293 K293 \text{ K}. Find its concentration (mass by mass percentage) at this temperature.

Solution:

Mass of solute (NaClNaCl) = 36 g36 \text{ g}. Mass of solvent (H2OH_2O) = 100 g100 \text{ g}. Mass of solution = Mass of solute+Mass of solvent=36 g+100 g=136 g\text{Mass of solute} + \text{Mass of solvent} = 36 \text{ g} + 100 \text{ g} = 136 \text{ g}. Concentration (%)=36136×10026.47%(\%) = \frac{36}{136} \times 100 \approx 26.47\%.

Explanation:

The concentration is calculated by dividing the mass of the solute by the total mass of the resulting solution (solute + solvent) and multiplying by 100100 to get the percentage.

Problem 2:

Identify the dispersed phase and dispersion medium in 'Fog'.

Solution:

Dispersed Phase: Liquid (H2OH_2O droplets). Dispersion Medium: Gas (Air).

Explanation:

Fog is an aerosol where tiny liquid water droplets are dispersed in a gaseous medium.

Problem 3:

Why does a beam of light become visible when passed through a colloidal solution of soap and water?

Solution:

This occurs due to the Tyndall Effect. The size of the soap particles (1 nm1 \text{ nm} to 100 nm100 \text{ nm}) is large enough to scatter the incident light rays in different directions, making the path of the beam visible.

Explanation:

True solutions do not show this effect because their particles are smaller than 1 nm1 \text{ nm} and cannot scatter light.

Suspensions and Colloids - Revision Notes & Key Formulas | CBSE Class 9 Science