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Chemistry: Matter, Mixtures, and Separation - Classification and Particulate Nature of Matter

Grade 8IB

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

🔑Concepts

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Matter is defined as anything that has mass and occupies space. It is composed of tiny, discrete particles (atoms, molecules, or ions) that are in constant motion.

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The Kinetic Molecular Theory states that the speed and spacing of particles depend on their energy. In a Solid, particles are closely packed in a regular lattice; in a Liquid, they are close but can flow; in a Gas, they are far apart and move randomly at high speeds.

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A Pure Substance consists of only one type of particle. This includes Elements (composed of one type of atom, like AuAu or O2O_2) and Compounds (different atoms chemically bonded in fixed ratios, like H2OH_2O or NaClNaCl).

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A Mixture contains two or more substances physically blended but not chemically joined. Homogeneous mixtures (solutions) have a uniform composition (e.g., salt dissolved in water), whereas Heterogeneous mixtures have distinct phases (e.g., sand in water).

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Phase Changes occur when thermal energy is added or removed, changing the KEKE (Kinetic Energy) of particles. For instance, sublimation is the direct change from solid to gas (S→GS \rightarrow G).

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Separation Techniques leverage physical properties: filtration for insoluble solids, evaporation/distillation for boiling point differences, and chromatography for solubility differences.

📐Formulae

Density(ρ)=Mass(m)Volume(V)\text{Density} (\rho) = \frac{\text{Mass} (m)}{\text{Volume} (V)}

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

Percentage Composition=(Mass of componentTotal mass of mixture)×100%\text{Percentage Composition} = \left( \frac{\text{Mass of component}}{\text{Total mass of mixture}} \right) \times 100\%

T(K)=T(∘C)+273.15T(K) = T(^\circ C) + 273.15

💡Examples

Problem 1:

A student dissolves 25 g25\text{ g} of sugar in 225 g225\text{ g} of water. Calculate the percentage by mass of the sugar in the resulting solution.

Solution:

Step 1: Total mass = 25 g+225 g=250 g25\text{ g} + 225\text{ g} = 250\text{ g}.
Step 2: Percentage=25250×100=10%\text{Percentage} = \frac{25}{250} \times 100 = 10\%.

Explanation:

The percentage concentration is found by dividing the mass of the solute by the total mass of the solution (solute + solvent) and multiplying by 100100.

Problem 2:

Calculate the density of a block of wood that has a mass of 600 g600\text{ g} and a volume of 750 cm3750\text{ cm}^3.

Solution:

ρ=600750=0.8 g/cm3\rho = \frac{600}{750} = 0.8\text{ g/cm}^3

Explanation:

Using the formula ρ=mV\rho = \frac{m}{V}, we divide mass by volume. Since the density is less than 1.0 g/cm31.0\text{ g/cm}^3 (the density of water), this wood will float.

Problem 3:

Determine the mass of a residue left if 500 g500\text{ g} of a salt solution is evaporated, given the solution was 15%15\% salt by mass.

Solution:

Mass of salt=500×15100=75 g\text{Mass of salt} = 500 \times \frac{15}{100} = 75\text{ g}

Explanation:

To find the mass of the solute from the percentage, multiply the total mass by the decimal form of the percentage (0.150.15).