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Exploring Mixtures and their Separation - Methods of Separation of Homogeneous Mixtures

Grade 9CBSE

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

🔑Concepts

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Homogeneous mixtures are those where the components are uniformly distributed, making it impossible to distinguish them by sight. Separation methods for these mixtures exploit differences in physical properties like boiling point and solubility.

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Evaporation: Used to separate a non-volatile solute (like dye) from a volatile solvent (like water). The solvent escapes as vapor, leaving the solute behind.

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Simple Distillation: Used to separate components of a mixture containing two miscible liquids that boil without decomposition and have a sufficient difference in their boiling points, typically more than 25 K25\text{ K} (or 25∘C25^\circ\text{C}).

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Fractional Distillation: Used when the difference in boiling points of miscible liquids is less than 25 K25\text{ K}. It employs a fractionating column packed with glass beads to provide a surface for vapors to cool and condense repeatedly. Example: Separating different gases from air or fractions from petroleum.

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Chromatography: A technique used for the separation of those solutes that dissolve in the same solvent. It is based on the principle that different components move at different speeds on the stationary phase (like filter paper).

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Crystallisation: A process that separates a pure solid in the form of its crystals from a solution. It is considered better than evaporation because some solids decompose or get charred on heating to dryness, and some impurities may remain in solution even after filtration.

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Separation of Gases from Air: Air is a homogeneous mixture. It is separated into its components (Oxygen, Argon, Nitrogen) by first compressing and cooling the air to form liquid air, followed by fractional distillation.

📐Formulae

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

ΔT=Tboiling, component 1−Tboiling, component 2\Delta T = T_{\text{boiling, component 1}} - T_{\text{boiling, component 2}}

💡Examples

Problem 1:

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

Solution:

Mass percentage=(50500)×100=10%\text{Mass percentage} = \left( \frac{50}{500} \right) \times 100 = 10\%

Explanation:

First, calculate the total mass of the solution: 50 (solute)+450 (solvent)500 (solution)\begin{array}{r} 50 \text{ (solute)} \\ + 450 \text{ (solvent)} \\ \hline 500 \text{ (solution)} \end{array} Then apply the formula: 50500×100\frac{50}{500} \times 100. The result is 10%10\%.

Problem 2:

Which method would you use to separate a mixture of Acetone (Boiling point 56∘C56^\circ\text{C}) and Water (Boiling point 100∘C100^\circ\text{C})?

Solution:

Simple Distillation.

Explanation:

Since the difference in boiling points is 100∘C−56∘C=44∘C100^\circ\text{C} - 56^\circ\text{C} = 44^\circ\text{C}, which is greater than 25∘C25^\circ\text{C} (25 K25\text{ K}), simple distillation is the most effective method.

Problem 3:

Explain why crystallisation is a better technique than simple evaporation for purifying a solid like Copper Sulphate.

Solution:

Crystallisation prevents decomposition of the solute.

Explanation:

During evaporation, the heat may cause some solids to decompose or char (like sugar). Also, impurities that are soluble might remain in the solute after evaporation, whereas in crystallisation, only the pure substance forms crystals, leaving impurities behind in the mother liquor.