SCIENCE · INVESTIGATE AND EXPLAIN

Distinguish a dissolved mixture from a pure substance

Compare total solute with the amount that can remain dissolved under fixed conditions.

Learning outcome: Distinguish a homogeneous solution from the whole heterogeneous sample and explain saturation without confusing it with dissolution rate.

Starting recap

A mixture contains more than one substance. A solution is homogeneous at the relevant scale; visible residue can coexist with a homogeneous liquid phase.

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Try

25 g of fictional solid A is mixed with 100 g water and 10 g insoluble B. The stipulated capacity is 20 g A per 100 g water.

Open the experiment

Predict

Predict which components remain solid and whether the clear liquid contains A.

Your prediction and reason

Experiment

Capture (15,100), (20,100), and (25,100). Distinguish below capacity, exactly at capacity and above capacity. Then increase water to 200 g with 25 g A.

Follow a mixture through dissolution and filtration

Worked teaching model · synthetic values · Fictional nonvolatile solid A dissolves up to 20 g per 100 g water at fixed conditions; solid B is insoluble and fixed at 10 g. Ideal complete equilibration and filtration, no retained liquid or transfer loss. Complete evaporation moves water to the surroundings without decomposing A. This is not a physical heating procedure.

Initial solid A: 25 g Water supplied: 100 g

Reset keeps your prediction, captures and practice history.

Follow the branches in order. The filtrate is an intermediate stream, not an additional final mass to add again. Filter residue, recovered A and water in the surroundings form the final inventory. Boxes show composition; their areas are not mass scales.

Calculated values on the same fixed scale
Row	Quantity	Value (g)
1	Initial total mass	135
2	Final total including surroundings	135
3	Final B in filter residue	10
4	Final A in filter residue	5
5	Final A recovered from filtrate	20
6	Final water in surroundings	100

20 g A dissolves; filter residue contains 5 g A and 10 g B.

Capacity = 0.20 × water mass = 20 g A. Filtrate mass before evaporation = 120 g. A recovered from filtrate = 80.0% of initial A. Final component rows are disjoint; do not add initial and final totals together. Dissolved A passes the ideal filter; water lost from the vessel remains in the surroundings.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Classify composition and phase separately

A pure substance has a definite chemical identity and composition; a mixture contains more than one substance. A compound is not simply another name for any mixture of elements. Water has a molecular composition, whereas water containing dissolved solute is a mixture whose proportions can vary. A sample can appear uniform without being a pure substance. Background: classification of matter, OpenStax Chemistry 2e.

A homogeneous mixture has uniform composition at the relevant scale. A heterogeneous sample contains distinguishable regions or phases. If undissolved solid lies beneath a uniform solution, the liquid phase can be homogeneous while the entire sample is heterogeneous. Identify which part of the sample your statement describes.

Use a stated solubility rather than assuming everything dissolves

The lab uses fictional solids A and B. At its fixed conditions, A can dissolve up to 20 g per 100 g water; B is insoluble and always starts at 10 g. These are programmed properties, not measurements of salt, sugar or another named material. Solubility concerns the equilibrium capacity under specified conditions. Background: solubility, OpenStax Chemistry 2e.

For 100 g water, adding 15 g A leaves all A dissolved and the liquid below capacity. Adding 20 g reaches capacity, with no excess A required to remain solid. Adding 25 g leaves 20 g dissolved and 5 g undissolved. B remains solid in all three cases, so the whole sample still contains a solid phase even when all A has dissolved.

Let w be water mass and a initial A mass, both in grams. Capacity is 0.20w grams of A. Dissolved A is the smaller of a and 0.20w; undissolved A is the difference between initial and dissolved A. These expressions account for each component instead of treating all visible solid as the same material.

Saturation and concentration answer different questions

Saturated means the solution has reached its equilibrium solubility for the specified solute and conditions. Concentrated is a relative description of how much solute is present. A substance with low solubility can have a saturated solution at a low concentration. The word saturated is not a synonym for a large mass of solid in the container.

The ratio here is grams A per grams of water. It is not molarity, and its denominator is not final solution mass. At saturation with 20 g A and 100 g water, the liquid contains 120 g total and its A mass fraction is 20/120, about 16.7%. Applying the ratio to the wrong denominator changes the result.

Transfer: 45 g A is supplied to 150 g water under the same stipulated conditions. Predict dissolved A, undissolved A and the composition of the liquid phase.

Check the capacity calculation

Capacity is 0.20 × 150 = 30 g A. Thirty grams dissolve and 15 g A remain solid. The liquid phase contains 30 g A plus 150 g water, totaling 180 g; the separate 10 g B is not part of that liquid total.

A final state does not describe how fast it forms

The calculation assumes equilibration has occurred. It has no time axis and cannot predict how long dissolving takes. Stirring or smaller particles can affect the approach to equilibrium without changing the fixed capacity used here. Temperature can affect actual solubility, but the model does not supply a temperature relationship and does not justify a universal rule that warming dissolves more of every solute.

Disappearance of visible solid does not mean its atoms disappeared. In this model the substance is dispersed in the liquid and can be recovered under the stated nonreaction assumptions. Real dissolution may involve dissociation or reaction depending on the system; observation and chemical evidence are needed before generalizing. Use the phase and component labels to explain your captures, rather than judging composition from appearance alone.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Explain why disappearance of visible A does not establish that it was destroyed or chemically converted. Separate the liquid-phase composition from the composition of the entire sample.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

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Previous: Decide what a coarse density measurement can establish Next: Design a separation and account for every fraction

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