krit.club logo

Is Matter Around Us Pure - Mixtures

Grade 9CBSE

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

🔑Concepts

A mixture consists of more than one kind of pure form of matter. Substances in a mixture can be separated by physical methods.

A Homogeneous mixture has a uniform composition throughout its mass, with no visible boundaries of separation between the constituents (e.g., a solution of SugarSugar in H2OH_{2}O).

A Heterogeneous mixture contains physically distinct parts and has a non-uniform composition (e.g., a mixture of SaltSalt and IronIron filings, or OilOil and WaterWater).

A Solution is a homogeneous mixture of two or more substances. The component that dissolves the other component is the Solvent, and the component being dissolved is the Solute.

Properties of a solution: Particle size is smaller than 1extnm1 ext{ nm} (109extm10^{-9} ext{ m}) in diameter, they do not scatter light, and they are stable (solute does not settle).

A Suspension is a heterogeneous mixture in which the solute particles do not dissolve but remain suspended throughout the bulk of the medium. Particles are visible to the naked eye (>100extnm> 100 ext{ nm}).

A Colloid is a heterogeneous mixture where particle size is intermediate between a true solution and a suspension (between 1extnm1 ext{ nm} and 1000extnm1000 ext{ nm}). They exhibit the Tyndall Effect, which is the scattering of a beam of light.

Solubility is the maximum amount of a solute that can be dissolved in a given amount of solvent at a specific temperature. A solution is Saturated when no more solute can be dissolved at that temperature.

📐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

Volume by volume percentage of a solution=(Volume of soluteVolume of solution)×100\text{Volume by volume percentage of a solution} = \left( \frac{\text{Volume of solute}}{\text{Volume of solution}} \right) \times 100

💡Examples

Problem 1:

A solution contains 40extg40 ext{ g} of common salt in 320extg320 ext{ g} of water. Calculate the concentration in terms of mass by mass percentage of the solution.

Solution:

Mass of solute (SaltSalt) = 40extg40 ext{ g}. Mass of solvent (WaterWater) = 320extg320 ext{ g}. Total mass of solution = 40extg+320extg=360extg40 ext{ g} + 320 ext{ g} = 360 ext{ g}. Concentration = 40360×100=11.11%\frac{40}{360} \times 100 = 11.11\%.

Explanation:

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

Problem 2:

To make a saturated solution, 36extg36 ext{ g} of sodium chloride is dissolved in 100extg100 ext{ g} of water at 293extK293 ext{ K}. Find its concentration at this temperature.

Solution:

Mass of solute (NaClNaCl) = 36extg36 ext{ g}. Mass of solvent (H2OH_{2}O) = 100extg100 ext{ g}. Mass of solution = 36extg+100extg=136extg36 ext{ g} + 100 ext{ g} = 136 ext{ g}. Concentration = (36136)×10026.47%\left( \frac{36}{136} \right) \times 100 \approx 26.47\%.

Explanation:

Concentration is expressed as the percentage of solute relative to the total solution mass (solute + solvent).

Mixtures - Revision Notes & Key Formulas | CBSE Class 9 Science