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The Amazing World of Solutes, Solvents, and Solutions - Types of Solutions and Solubility

Grade 7CBSE

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

🔑Concepts

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A Solution is a homogeneous mixture of two or more substances. The component that is dissolved is the Solute, and the medium in which it is dissolved is the Solvent.

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Water (H2OH_2O) is often called the Universal Solvent because it can dissolve a wide variety of substances.

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An Unsaturated Solution is one in which more solute can be dissolved at a given temperature without increasing the temperature.

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A Saturated Solution is a solution in which no more solute can be dissolved at a specific temperature. Any additional solute added will settle at the bottom.

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Solubility is the maximum amount of solute (usually in grams) that can be dissolved in 100 g100\text{ g} of a solvent at a specific temperature.

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Factors affecting solubility: For solid solutes in liquid solvents, solubility generally increases with an increase in Temperature. For gases in liquids, solubility decreases as temperature increases.

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A Supersaturated Solution contains more solute than a saturated solution at the same temperature, usually prepared by cooling a saturated solution made at a higher temperature.

📐Formulae

Mass of Solution=Mass of Solute+Mass of Solvent\text{Mass of Solution} = \text{Mass of Solute} + \text{Mass of Solvent}

Concentration (Mass by Mass %)=(Mass of SoluteMass of Solution)×100\text{Concentration (Mass by Mass \%)} = \left( \frac{\text{Mass of Solute}}{\text{Mass of Solution}} \right) \times 100

Solubility=Mass of Solute in a saturated solutionMass of Solvent×100\text{Solubility} = \frac{\text{Mass of Solute in a saturated solution}}{\text{Mass of Solvent}} \times 100

💡Examples

Problem 1:

A solution is prepared by dissolving 40 g40\text{ g} of common salt (NaClNaCl) in 320 g320\text{ g} of water. Calculate the concentration in terms of mass by mass percentage of the solution.

Solution:

Mass of solute (NaClNaCl) = 40 g40\text{ g}. Mass of solvent (water) = 320 g320\text{ g}. Total mass of solution = 40 g+320 g=360 g40\text{ g} + 320\text{ g} = 360\text{ g}. Concentration = (40360)×100=11.11%\left( \frac{40}{360} \right) \times 100 = 11.11\%.

Explanation:

To find the mass percentage, we divide the mass of the solute by the total mass of the solution (sum of solute and solvent) and multiply by 100100.

Problem 2:

If the solubility of Potassium Nitrate (KNO3KNO_3) is 32 g32\text{ g} at 20∘C20^\circ\text{C}, how much solute is needed to saturate 50 g50\text{ g} of water at the same temperature?

Solution:

Solubility is 32 g32\text{ g} per 100 g100\text{ g} of water. For 50 g50\text{ g} of water, the amount of KNO3KNO_3 required = 32100×50=16 g\frac{32}{100} \times 50 = 16\text{ g}.

Explanation:

Since solubility is defined per 100 g100\text{ g} of solvent, if the amount of solvent is halved (from 100 g100\text{ g} to 50 g50\text{ g}), the amount of solute required to reach saturation is also halved.

Problem 3:

What happens to the solubility of sugar in water if the temperature is increased from 25∘C25^\circ\text{C} to 70∘C70^\circ\text{C}?

Solution:

The solubility of sugar increases significantly.

Explanation:

In most solid-liquid solutions, kinetic energy increases with temperature, allowing solvent molecules to break apart solute particles more effectively, thus increasing solubility.