Classifying Matter & Separating Mixtures — Practice · Grade 11 Science

Chapter: Investigating Matter

What this practice covers

By the end of Classifying Matter & Separating Mixtures you should be able to:

8 questions, written answer and numeric, each with a worked explanation.

Vocabulary used: pure substance, mixture, filtrate.

Problem 1

Why does a coffee filter not remove dissolved sugar?
Answer and explanation

Dissolved molecules pass through its pores with the liquid; filtration removes larger particles.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 2

Explain why boiling and electrolysis of water are different kinds of change.
Answer and explanation

Boiling changes phase without changing H₂O; electrolysis is a chemical decomposition into hydrogen and oxygen.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 3

A mixture contains 5 g salt and 8 g sand. After complete physical separation, what total dry solid mass should be recovered in a closed no-loss model?
Hints

List the given quantities, select a relationship, and check units before substituting.

Answer and explanation

13 g

5+8=13; material is redistributed, not created.

Problem 4

A wet towel loses 30 g while drying. Explain conservation.
Answer and explanation

Water crosses the towel-system boundary into air; include that vapor to conserve the total mass.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 5

Classify pure copper, distilled water and air as element, compound or mixture. Show the reasoning, not just the conclusion.
Hints

Filtering alone cannot normally remove dissolved salt ions.

Answer and explanation

Copper is an element, water a compound, and air a mixture of gases under ordinary conditions.

Problem 6

Why does a variable composition support calling a material a mixture?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Mixture proportions can vary without a single fixed chemical formula; a pure compound has defined composition.

Problem 7

Choose a separation step because a physical property differsChoose a separation step because a physical property differsSalt + sand mixtureWater dissolves the saltFilter: sand retainedEvaporate filtrate → saltFiltration separates phases, not dissolved ions from water.
Original conceptual sand/salt route; actual material quantities, apparatus and drying conditions require a specified procedure.
Interpret the supplied visual. Why does filtering salty water not normally produce salt-free water?
Answer and explanation

Dissolved ions travel with water through an ordinary filter. A different separation process must exploit volatility, membranes or other relevant properties.

A separation method uses a property difference without requiring a chemical identity change. Filtration retains particles larger than a filter’s effective pores but does not normally remove dissolved ions. Selective dissolution can move salt into water while sand remains as a solid phase. Evaporation recovers a nonvolatile solute, while distillation can also collect condensed solvent vapor. Evaluate solubility, volatility, particle size and stability before choosing a method; a sequence can need several complementary steps.

Dissolved ions travel with water through an ordinary filter. A different separation process must exploit volatility, membranes or other relevant properties.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 8

A mixture contains 6.00 g salt and 4.00 g sand. You recover 5.10 g of dry salt-containing material. Calculate apparent salt recovery and explain why its purity remains uncertain.
Answer and explanation

Apparent recovery is 5.10/6.00 ×100 = 85.0%. The calculation treats the recovered mass as salt, but residual sand or other substances require a separate purity test. Distinguish mass loss from composition evidence.

Apparent recovery is 5.10/6.00 ×100 = 85.0%. The calculation treats the recovered mass as salt, but residual sand or other substances require a separate purity test. Distinguish mass loss from composition evidence.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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