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Nature of Matter: Elements, Compounds, and Mixtures - What Are 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 physically in any proportion, without any chemical reaction taking place.

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The components of a mixture retain their individual chemical and physical properties. For example, in a mixture of Iron (FeFe) and Sulfur (SS), the iron remains magnetic.

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Mixtures are classified into two types: Homogeneous and Heterogeneous.

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Homogeneous Mixtures have a uniform composition throughout. The particles of the components are not visible even under a microscope. Examples include air (N2,O2,Ar,CO2N_2, O_2, Ar, CO_2), alloys like Brass (Cu+ZnCu + Zn), and salt solutions.

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Heterogeneous Mixtures have a non-uniform composition. The components can often be seen with the naked eye or under a microscope. Examples include soil, a mixture of oil and water, and chalk powder in water.

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The components of a mixture can be separated by physical methods such as filtration, evaporation, distillation, sublimation, and magnetic separation.

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A solution is a homogeneous mixture of two or more substances. It consists of a solute (substance being dissolved) and a solvent (substance in which the solute is dissolved).

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Concentration of a solution is the amount of solute present in a given amount (mass or volume) of solution.

📐Formulae

Mass percentage of a solution=(Mass of soluteMass of solution)×100\text{Mass percentage of a solution} = \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}

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

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

💡Examples

Problem 1:

A solution contains 50 g50\text{ g} of common salt in 450 g450\text{ g} of water. Calculate the concentration in terms of mass by mass percentage of the solution.

Solution:

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

Explanation:

To find the mass percentage, we first calculate the total mass of the solution by adding the mass of the solute and the solvent. Then, we apply the mass percentage formula.

Problem 2:

Identify whether a mixture of sand and water is homogeneous or heterogeneous, and suggest a method to separate them.

Solution:

The mixture is Heterogeneous. It can be separated using Filtration.

Explanation:

Sand does not dissolve in water and its particles remain suspended, making the composition non-uniform. Since the sand particles are relatively large and insoluble, they can be trapped by filter paper while the water passes through.

Problem 3:

Calculate the mass of Glucose (C6H12O6C_6H_{12}O_6) needed to prepare 250 g250\text{ g} of a 5%5\% (mass/mass) solution.

Solution:

Mass percentage=5%Mass of solution=250 g5=(Mass of solute250)×100Mass of solute=5×250100=12.5 g\begin{array}{r} \text{Mass percentage} = 5\% \\ \text{Mass of solution} = 250\text{ g} \\ 5 = \left( \frac{\text{Mass of solute}}{250} \right) \times 100 \\ \text{Mass of solute} = \frac{5 \times 250}{100} = 12.5\text{ g} \end{array}

Explanation:

By rearranging the mass percentage formula: Mass of solute=Percentage×Mass of solution100\text{Mass of solute} = \frac{\text{Percentage} \times \text{Mass of solution}}{100}, we can find the required quantity of glucose.