krit.club logo

The Amazing World of Solutes, Solvents, and Solutions - Understanding Solubility and Saturated Solutions

Grade 8CBSE

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

🔑Concepts

•

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). For example, in a sugar solution, sugar is the solute and water (H2OH_2O) is the solvent.

•

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

•

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

•

The solubility of most solid solutes in liquid solvents increases with an increase in temperature. Conversely, cooling a saturated solution often leads to the precipitation of some dissolved solute.

•

An unsaturated solution is a solution in which more quantity of solute can be dissolved without raising its temperature.

•

Concentration of a solution is the amount of solute present in a given quantity of the solution. It is commonly expressed as a mass percentage or volume percentage.

📐Formulae

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

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 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

💡Examples

Problem 1:

A solution is prepared by dissolving 50 g50 \text{ g} of common salt (NaClNaCl) in 450 g450 \text{ g} of water (H2OH_2O). Calculate the concentration of the solution in terms of mass by mass percentage.

Solution:

Mass of solute (NaClNaCl) = 50 g50 \text{ g} \ Mass of solvent (H2OH_2O) = 450 g450 \text{ g} \ Mass of solution = 50 g+450 g=500 g50 \text{ g} + 450 \text{ g} = 500 \text{ g} \ Concentration = (50500)×100=10%\left( \frac{50}{500} \right) \times 100 = 10\%

Explanation:

To find the mass percentage, we first determine the total mass of the solution by adding the solute and solvent. Then, we divide the mass of the solute by the total mass and multiply by 100.

Problem 2:

The solubility of potassium nitrate (KNO3KNO_3) is 62 g62 \text{ g} in 100 g100 \text{ g} of water at 313 K313 \text{ K}. What mass of KNO3KNO_3 would be needed to produce a saturated solution of KNO3KNO_3 in 50 g50 \text{ g} of water at the same temperature?

Solution:

Solubility in 100 g100 \text{ g} water = 62 g62 \text{ g} \ For 1 g1 \text{ g} of water, mass needed = 62100 g\frac{62}{100} \text{ g} \ For 50 g50 \text{ g} of water, mass needed = 62100×50=31 g\frac{62}{100} \times 50 = 31 \text{ g}

Explanation:

Since solubility is defined per 100 g100 \text{ g} of solvent, we use a simple proportion to find the amount needed for 50 g50 \text{ g} of water at the same temperature.