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Exploring Mixtures and their Separation - How Can We Classify Mixtures?

Grade 9CBSE

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

🔑Concepts

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A mixture is a substance that consists of two or more types of particles (atoms or molecules) which are not chemically combined but are physically mixed together in any proportion.

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Homogeneous Mixtures have a uniform composition throughout. The components are not visible separately. Example: A solution of sugar in water.

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Heterogeneous Mixtures have a non-uniform composition and contain physically distinct parts. Example: A mixture of salt and iron filings.

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A Solution is a homogeneous mixture of two or more substances. The particle size is less than 1 nm1\text{ nm} (10−9 m10^{-9}\text{ m}) in diameter. They do not scatter a beam of light passing through them.

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A Suspension is a heterogeneous mixture in which the solute particles do not dissolve but remain suspended throughout the bulk of the medium. Particle size is larger than 100 nm100\text{ nm}.

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A Colloid is a heterogeneous mixture where the particle size (1 nm1\text{ nm} to 100 nm100\text{ nm}) is intermediate between a true solution and a suspension. They are stable and show the Tyndall effect (scattering of a beam of light).

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Saturated Solution: A solution in which no more solute can be dissolved at a given temperature.

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Solubility: The maximum amount of solute that can be dissolved in 100 g100\text{ g} of a solvent at a specified temperature.

📐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}

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

💡Examples

Problem 1:

A solution contains 50 g50\text{ g} of sugar dissolved in 350 g350\text{ g} of water. Calculate the concentration in terms of mass by mass percentage of the solution.

Solution:

Mass of solute (sugar)=50 g\text{Mass of solute (sugar)} = 50\text{ g} Mass of solvent (water)=350 g\text{Mass of solvent (water)} = 350\text{ g} Mass of solution=50 g+350 g=400 g\text{Mass of solution} = 50\text{ g} + 350\text{ g} = 400\text{ g} Concentration=50400×100=12.5%\text{Concentration} = \frac{50}{400} \times 100 = 12.5\%

Explanation:

First, we find the total mass of the solution by adding the mass of the solute and the solvent. Then, we apply the mass by mass percentage formula to find the concentration.

Problem 2:

Calculate the mass of potassium sulfate required to prepare its 10%10\% (mass/mass) solution in 180 g180\text{ g} of water.

Solution:

Let the mass of solute be x gx\text{ g}. Mass of solution=(x+180) g\text{Mass of solution} = (x + 180)\text{ g} Concentration=xx+180×100=10\text{Concentration} = \frac{x}{x + 180} \times 100 = 10 100x=10(x+180)100x = 10(x + 180) 100x=10x+1800100x = 10x + 1800 90x=180090x = 1800 x=20 gx = 20\text{ g}

Explanation:

To find the mass of the solute when the solvent mass and concentration are given, we set up an algebraic equation based on the mass by mass percentage formula.