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The Amazing World of Solutes, Solvents, and Solutions - Saturated and Unsaturated Solutions

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. It consists of a solute (the substance being dissolved) and a solvent (the component that dissolves the solute, often H2OH_2O).

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An Unsaturated Solution is a solution in which more amount of solute can be dissolved in the solvent at a given 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 simply settle at the bottom.

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

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The solubility of most solids in liquids increases with an increase in temperature (TT). Therefore, a saturated solution may become unsaturated if the temperature is raised.

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If a saturated solution is cooled, the solubility decreases, and some of the dissolved solute may separate out in the form of crystals.

📐Formulae

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

Concentration of Solution (Mass/Mass %)=(Mass of SoluteMass of Solution)×100\text{Concentration of Solution (Mass/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 30 g30\text{ g} of sugar in 170 g170\text{ g} of water. Calculate the concentration of the solution in terms of mass by mass percentage.

Solution:

Mass of solute (sugar) = 30 g30\text{ g}. Mass of solvent (water) = 170 g170\text{ g}. Therefore, Mass of solution = 30 g+170 g=200 g30\text{ g} + 170\text{ g} = 200\text{ g}. Concentration = 30200×100=15%\frac{30}{200} \times 100 = 15\%.

Explanation:

To find the concentration, we must use the total mass of the solution (solute + solvent) as the denominator.

Problem 2:

If the solubility of Potassium Nitrate (KNO3KNO_3) is 32 g32\text{ g} at 20∘C20^\circ\text{C} in 100 g100\text{ g} of water, what happens if we add 40 g40\text{ g} of KNO3KNO_3 to 100 g100\text{ g} of water at this temperature?

Solution:

The solution will become saturated after 32 g32\text{ g} is dissolved. The remaining 8 g8\text{ g} (40 g−32 g40\text{ g} - 32\text{ g}) will remain undissolved at the bottom.

Explanation:

Since the maximum capacity (solubility) is 32 g32\text{ g}, the solvent cannot hold any more solute at that specific temperature.

Problem 3:

A saturated solution of common salt (NaClNaCl) is prepared at 80∘C80^\circ\text{C}. What will you observe if the solution is allowed to cool down to room temperature (25∘C25^\circ\text{C})?

Solution:

You will observe crystals of salt forming at the bottom of the container.

Explanation:

Solubility is directly proportional to temperature for most solids. As temperature decreases (T↓T \downarrow), the solubility decreases, and the excess solute that was dissolved at the higher temperature precipitates out.