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The Amazing World of Solutes, Solvents, and Solutions - How does temperature affect the solubility of a solid in a liquid?

Grade 8CBSE

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. It consists of a solute (the substance being dissolved) and a solvent (the medium in which the solute is dissolved).

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

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A Saturated Solution is a solution in which no more solute can be dissolved at a given temperature. If more solute is added, it will simply settle at the bottom.

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An Unsaturated Solution is a solution that contains less than the maximum amount of solute that can be dissolved at that temperature.

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Effect of Temperature on Solubility: For most solids in liquids, solubility increases with an increase in temperature. This is because higher temperature provides more kinetic energy to the solvent molecules, allowing them to break the intermolecular forces of the solute more effectively.

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If a saturated solution at a high temperature is cooled, the solubility decreases, and the excess solute typically separates out in the form of crystals.

📐Formulae

Concentration of solution (Mass/Mass %)=Mass of soluteMass of solution×100\text{Concentration of solution (Mass/Mass \%)} = \frac{\text{Mass of solute}}{\text{Mass 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}

Solubility=Mass of solute (in g)Mass of solvent (in g)×100\text{Solubility} = \frac{\text{Mass of solute (in g)}}{\text{Mass of solvent (in g)}} \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:

  1. Identify Mass of Solute = 50 g50\text{ g}.
  2. Identify Mass of Solvent = 450 g450\text{ g}.
  3. Calculate Mass of Solution: 50+450500\begin{array}{r} 50 \\ + 450 \\ \hline 500 \end{array} Mass of solution = 500 g500\text{ g}.
  4. Apply formula: Concentration=50500×100\text{Concentration} = \frac{50}{500} \times 100.

Explanation:

The concentration is calculated by dividing the mass of the solute by the total mass of the solution (solute + solvent) and multiplying by 100100. 50500×100=10%\frac{50}{500} \times 100 = 10\%.

Problem 2:

If the solubility of Potassium Nitrate is 62 g62\text{ g} at 313 K313\text{ K}, what mass of Potassium Nitrate would be needed to produce a saturated solution in 50 g50\text{ g} of water at the same temperature?

Solution:

Given: Solubility in 100 g100\text{ g} water = 62 g62\text{ g}. Mass of water provided = 50 g50\text{ g}. Mass needed = 62100×50\frac{62}{100} \times 50.

Explanation:

Since solubility is measured per 100 g100\text{ g} of solvent, for 50 g50\text{ g} of water (which is half of 100 g100\text{ g}), we need half the amount of solute. 622=31 g\frac{62}{2} = 31\text{ g}.