Answer key · Unit 6

Solutions.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 4

Lesson 1 of 4 · About 20 minutes

Dissolving

What happens to salt that vanishes in water: solvent and solute, ions wrapped in water molecules, mass that is conserved, and what sets a solution apart from any mixture.

Open the lesson

The 4 steps and their answers

Step 1 of 4
Open this step
Where it went
Practice · Choose
Question
10 g of salt went into 100 g of water, and a minute later you cannot see it. Where is the salt?

Correct answer
- Correct Spread all through the water as single ions, too small to see
Setup, choices and feedback
The Scene at this step
Salt added
10 g
Water temperature
293 K
Stir
offdefault
Beaker / Particles
Beaker
Choices
Shown in a random order.
- Correct Spread all through the water as single ions, too small to see
- Wrong It has disappeared
  Misconception: Thinks a dissolved substance has disappeared, so its mass is lost.
  Feedback from the catalogue
  Dissolved salt has not vanished: its ions, $\mathrm{Na^+}$ ⁠ and $\mathrm{Cl^-}$ ⁠, are spread out between the water molecules, too small to see. Mass is conserved: 100 g of water and 20 g of salt weigh 120 g after the salt has dissolved too. And if you evaporate the water, the salt comes back.
- Wrong It melted into a liquid
  Misconception: Calls dissolving melting.
  Feedback from the catalogue
  Melting is one substance going from solid to liquid with heat, and salt melts only at 801 °C. Dissolving mixes two substances: the water pulls the ions off the crystal and surrounds each one. It happens in cold water too.
- Wrong It sank to the bottom as a fine powder
  Feedback
  A powder would cloud the water and settle slowly. Salt water is perfectly clear, and every drop of it is just as salty.
Feedback for a right answer
The salt has dissolved: water molecules pulled the ions off the crystal one by one, and they are now spread through all the water. You can prove it is there: taste it, or evaporate the water and the salt comes back.

Worked solution
1. Salt is an ionic compound: a crystal of $\mathrm{Na^+}$ ⁠ and $\mathrm{Cl^-}$ ⁠ ions.
2. In water each ion is pulled by the water molecules and comes away from the crystal.
3. The ions spread through all the liquid, and the solution stays clear and uniform.
The salt did not melt: salt melts only at 801 °C.
Step 2 of 4
Open this step
Inside the water
Practice · Predict, then look
Question
Look inside will show what is in the water. What will you see?

Correct answer
- Correct Separate $\mathrm{Na^+}$ ⁠ and $\mathrm{Cl^-}$ ⁠ ions, each surrounded by water molecules
Setup, choices and feedback
The Scene at this step
Salt added
10 g
Water temperature
293 K
Stir
offdefault
Beaker / Particles
Beakerfixed
Choices
- Correct Separate $\mathrm{Na^+}$ ⁠ and $\mathrm{Cl^-}$ ⁠ ions, each surrounded by water molecules
- Wrong Tiny grains of salt floating in the water
  Feedback
  A tiny grain is still a crystal of billions of ions, and it would scatter light and cloud the water. In a solution the crystal comes apart down to the last ion.
- Wrong Whole $\mathrm{NaCl}$ ⁠ molecules
  Feedback
  Salt has no molecules: it is a lattice of ions. In water the sodium and the chlorine come apart.
- Wrong Only water
  Misconception: Thinks a dissolved substance has disappeared, so its mass is lost.
  Feedback from the catalogue
  Dissolved salt has not vanished: its ions, $\mathrm{Na^+}$ ⁠ and $\mathrm{Cl^-}$ ⁠, are spread out between the water molecules, too small to see. Mass is conserved: 100 g of water and 20 g of salt weigh 120 g after the salt has dissolved too. And if you evaporate the water, the salt comes back.
Once the prediction is locked in, the Scene changes
Beaker / Particles
Particlesfixed
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Each ion is wrapped in a shell of water molecules: the slightly negative oxygen side faces $\mathrm{Na^+}$ ⁠, and the slightly positive hydrogens face $\mathrm{Cl^-}$ ⁠. The shell is what holds the ions in solution.

Worked solution
1. A water molecule is polar: the oxygen is slightly negative and the hydrogens slightly positive.
2. The pull between these charges and the ions is strong enough to tear them off the crystal.
3. So in solution each ion is on its own, with water molecules arranged round it.
Step 3 of 4
Open this step
Mass is conserved
Practice · Calculate
Question
A glass on a balance holds 100 g of water. You add 20 g of salt and wait until it has all dissolved. What is the mass of the solution?

Correct answer
120 g ± 0.5 g
Setup, choices and feedback
The Scene at this step
Salt added
20 g
Water temperature
293 K
Stir
offdefault
Beaker / Particles
Beaker
Accepted within
± 0.5 g
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 100 g ± 0.5 g
  Misconception: Thinks a dissolved substance has disappeared, so its mass is lost.
  Feedback from the catalogue
  Dissolved salt has not vanished: its ions, $\mathrm{Na^+}$ ⁠ and $\mathrm{Cl^-}$ ⁠, are spread out between the water molecules, too small to see. Mass is conserved: 100 g of water and 20 g of salt weigh 120 g after the salt has dissolved too. And if you evaporate the water, the salt comes back.
Feedback for a right answer
120 g: the mass of the water plus the mass of the salt. In dissolving no particle is made and none disappears, so mass is conserved.

Worked solution
1. Mass of the solvent: 100 g. Mass of the solute: 20 g.
2. All the ions of the salt are still in the glass, only spread out.
3. Mass of the solution: $100 + 20 = 120$ ⁠.
4. The answer: 120 g.
If you answered 100 g, you took the dissolved salt to weigh nothing.
Step 4 of 4
Open this step
Solution or mixture
Practice · Choose
Question
Which of these mixtures is a solution?

Correct answer
- Correct Sugar in water, after stirring
Setup, choices and feedback
The Scene at this step
Salt added
10 g
Water temperature
293 K
Stir
offdefault
Beaker / Particles
Particles
Choices
Shown in a random order.
- Correct Sugar in water, after stirring
- Wrong Sand in water
  Feedback
  Sand does not dissolve: the grains settle, and you can filter them out. It is a mixture, but not a solution.
- Wrong Oil and water, after shaking
  Feedback
  The oil stays as separate drops, and within minutes forms its own layer. A solution stays uniform for good.
- Wrong Mud
  Feedback
  Mud is cloudy: solid bits of soil scattered in the water, settling bit by bit.
Feedback for a right answer
A solution is a uniform mixture in which the solute is spread through the solvent down to single molecules or ions: it is clear, it does not settle, and it passes through a filter. Sugar in water fits all of these.

Worked solution
1. In a solution the solute breaks into single particles within the solvent.
2. So a solution is clear, uniform, and does not separate over time.
3. Sand and mud settle, and oil forms a layer: they are non-uniform mixtures.
4. Only the sugar dissolves: its molecules spread out among the water molecules.

Lesson 2 of 4 · About 25 minutes

Solubility

How much salt water can hold: saturated solutions, solubility in grams per 100 grams of water, what stirring and heat do, and why gases dissolve less in warm water.

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The 5 steps and their answers

Step 1 of 5
Open this step
Saturated
Practice · Calculate
Question
In the Scene, 100 g of water at 20 °C holds at most 36.4 g of salt. 50 g has been added. How many grams of salt stay on the bottom?

Correct answer
13.6 g ± 0.1 g
Setup, choices and feedback
The Scene at this step
Salt added
50 gfixed
Water temperature
293 Kfixed
Stir
offdefault
Beaker / Particles
Beaker
Accepted within
± 0.1 g
Computed from the Scene at this step’s setting
Solute left as a pile on the bottom/nodes/beaker/outputs/undissolved
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0 g ± 0.1 g
  Misconception: Thinks any amount of solid will dissolve if you wait or stir long enough.
  Feedback from the catalogue
  There is a limit to how much the water can hold, the solubility: at 20 °C, 100 g of water holds at most about 36 g of salt. Beyond that the solution is saturated, and every extra gram stays on the bottom, however long you wait or stir.
Feedback for a right answer
$50 - 36.4 = 13.6$ ⁠ grams. The solution is saturated: it holds all it can, and the rest stays as crystals. The pile in the Scene will not go, even if you wait an hour.

Worked solution
1. The solubility at 20 °C: 36.4 g per 100 g of water.
2. We added 50 g, more than the solubility, so only 36.4 g dissolves.
3. Left over: $50 - 36.4 = 13.6$ ⁠.
4. The answer: 13.6 g.
Step 2 of 5
Open this step
Stirring
Practice · Predict, then look
Question
There is a pile of undissolved salt on the bottom. What will happen to it when you turn on Stir?

Correct answer
- Correct It will stay the same size
Setup, choices and feedback
The Scene at this step
Salt added
50 gfixed
Water temperature
293 Kfixed
Stir
offfixed
Beaker / Particles
Beaker
Choices
- Correct It will stay the same size
- Wrong It will dissolve, slowly or quickly
  Misconception: Thinks stirring raises how much can dissolve, not just how fast.
  Feedback from the catalogue
  Stirring speeds dissolving up: it carries the water that is already full of salt away from the crystals and brings fresh water. But it does not change the solubility. In a saturated solution, stirring will not dissolve one more gram.
- Wrong It will grow, because stirring pulls salt out of the water
  Feedback
  Stirring does not change how much salt the water holds, up or down.
Once the prediction is locked in, the Scene changes
Stir
onfixed
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The pile stays. Stirring speeds dissolving up while the solution is not yet saturated, but it does not raise the solubility. Only a change of temperature, or more water, can dissolve more.

Worked solution
1. The water already holds all the salt it can: the solution is saturated.
2. Stirring brings fresh water to the crystals, which helps only while some of the water can still take more salt.
3. Here none can, so the pile does not change.
Step 3 of 5
Open this step
Hot water
Practice · Predict, then look
Question
At 20 °C, 36.4 g of the 50 g dissolved. How many grams of salt will dissolve when you heat the water to 100 °C?

Correct answer
39.2 g ± 1 g
A number prediction counts as right within ± 1 g
Setup, choices and feedback
The Scene at this step
Salt added
50 gfixed
Water temperature
293 Kfixed
Stir
offdefault
Beaker / Particles
Beaker
Once the prediction is locked in, the Scene changes
Water temperature
373 Kfixed
The Scene then shows: 39.2 g · Solute dissolved in the 100 g of water/nodes/beaker/outputs/dissolved
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Only 39.2 g: less than three grams more. Table salt is an exception: its solubility hardly depends on temperature. Sugar, by contrast, dissolves two and a half times as much in boiling water as in cold.

Worked solution
1. The solubility of $\mathrm{NaCl}$ ⁠: 35.7 g at 0 °C and 39.2 g at 100 °C per 100 g of water.
2. So at 100 °C, 39.2 g dissolves, and 10.8 g is left.
If you guessed it would all dissolve, you assumed heat raises every substance’s solubility a lot. For most solids it does, but not for table salt.
Step 4 of 5
Open this step
Heat until it all dissolves
Practice · Reach the target
Question
The glass holds 38 g of salt, 1.6 g of it on the bottom. Salt added is fixed. Change Water temperature until all the salt has dissolved, then press Check.

Goal
- 0 kg ± 0.00001 kg
  At the start of the step: 0.0016 kg
Setup, choices and feedback
The Scene at this step
Salt added
38 gfixed
Water temperature
293 K
Stir
offfixed
Beaker / Particles
Beaker
Read from the Scene at
Solute left as a pile on the bottom/nodes/beaker/outputs/undissolved
One setting that reaches it
Water temperature
339 K
Feedback for a right answer
Above about 339 K, 66 °C, all 38 g dissolves. It took nearly 50 degrees of heating to dissolve just 1.6 g more.

Feedback for a wrong answer
Not yet. Heat it more: $35.7 + 0.035\,(T - 273)$ ⁠ has to reach 38.

Hint
The solubility rises by only 0.035 g per degree.

Worked solution
1. The solubility in the Scene: $35.7 + 0.035\,(T - 273)$ ⁠ grams per 100 g of water.
2. You want it to be at least 38 g: $0.035\,(T - 273) \ge 2.3$ ⁠.
3. $T - 273 \ge 66$ ⁠, that is $T \ge 339\ \mathrm{K}$ ⁠.
4. Set Water temperature to 340 K or more.
Step 5 of 5
Open this step
Gases
Practice · Choose
Question
A bottle of fizzy drink left in the sun loses its fizz faster than one from the fridge. Why?

Correct answer
- Correct Gases dissolve less in warm water
Setup, choices and feedback
The Scene at this step
Salt added
10 g
Water temperature
293 K
Stir
offdefault
Beaker / Particles
Particles
Choices
Shown in a random order.
- Correct Gases dissolve less in warm water
- Wrong In warm water everything dissolves more, so the gas is swallowed by the liquid
  Misconception: Thinks hotter water always dissolves much more of anything.
  Feedback from the catalogue
  It depends on the substance. The solubility of most solids rises with temperature, but table salt’s hardly does: 36 g at 20 °C and 39 g at 100 °C. And the solubility of gases falls when it is hot: that is why a warm fizzy drink goes flat fast.
- Wrong The drink starts to boil
  Feedback
  At 35 °C water is far from boiling. The bubbles are $\mathrm{CO_2}$ ⁠ leaving the solution, not water vapour.
Feedback for a right answer
A gas’s solubility falls with temperature: fast gas molecules escape the liquid more easily. Cold water holds nearly twice as much $\mathrm{CO_2}$ ⁠ as warm water, so a cold drink keeps its fizz.

Worked solution
1. The bubbles in the drink are carbon dioxide that was dissolved in it.
2. When warm the molecules move faster, and the gas molecules escape the liquid more easily.
3. So less gas stays dissolved in a warm drink, and the rest comes out as bubbles.

Lesson 3 of 4 · About 30 minutes

Concentration

How much solute is in each litre: concentration in grams per litre and moles per litre, how to make up a solution of a given concentration, and why the salt on the bottom does not count.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Grams per litre
Practice · Calculate
Question
The glass holds 10 g of salt dissolved in 100 mL of water. What is the concentration in grams per litre? Take the solution’s volume as 100 mL.

Correct answer
100 ± 1%
Setup, choices and feedback
The Scene at this step
Salt added
10 gfixed
Water temperature
293 K
Stir
offdefault
Beaker / Particles
Beaker
Accepted within
± 1%
A fixed value in the question
Wrong answers it recognises
- 0.1 ± 1%
  Misconception: Divides by the volume in millilitres when the concentration is per litre.
  Feedback from the catalogue
  Concentration is measured per litre. 100 mL is 0.1 L, so divide by 0.1, not by 100. Dividing by 100 gives an answer a thousand times too small.
Feedback for a right answer
$\dfrac{10\ \mathrm{g}}{0.1\ \mathrm{L}} = 100\ \mathrm{g/L}$ ⁠. A litre of this solution would hold 100 g of salt. Seawater holds about 35 g/L.

Hint
100 mL = 0.1 L.

Worked solution
1. Concentration in grams per litre: $c = \dfrac{m}{V}$ ⁠.
2. $m = 10\ \mathrm{g}$ ⁠, $V = 100\ \mathrm{mL} = 0.1\ \mathrm{L}$ ⁠.
3. $c = \dfrac{10}{0.1} = 100$ ⁠.
4. The answer: 100 g/L.
If you got 0.1, you divided by 100 millilitres: that is grams per millilitre.
Step 2 of 5
Open this step
Moles per litre
Practice · Calculate
Question
The glass holds 11.7 g of $\mathrm{NaCl}$ ⁠ (molar mass 58.44 g/mol) dissolved in 100 mL. What is the concentration?

Correct answer
2.002 mol/L ± 2%
Setup, choices and feedback
The Scene at this step
Salt added
11.7 gfixed
Water temperature
293 K
Stir
offdefault
Beaker / Particles
Beaker
Accepted within
± 2%
Computed from the Scene at this step’s setting
Dissolved solute: its moles over the 0.1 L (10⁻⁴ m³) of water/nodes/beaker/outputs/concentration
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.2 mol/L ± 2%
  Misconception: Divides by the volume in millilitres when the concentration is per litre.
  Feedback from the catalogue
  Concentration is measured per litre. 100 mL is 0.1 L, so divide by 0.1, not by 100. Dividing by 100 gives an answer a thousand times too small.
- 117 mol/L ± 2%
  Misconception: Divides grams by litres and calls the result mol/L.
  Feedback from the catalogue
  Molar concentration is moles per litre, $c = \dfrac{n}{V}$ ⁠. First turn grams into moles with the molar mass, $n = \dfrac{m}{M}$ ⁠; grams per litre is a different unit. For $\mathrm{NaCl}$ ⁠, $M = 58.44\ \mathrm{g/mol}$ ⁠.
Feedback for a right answer
$n = \dfrac{11.7}{58.44} = 0.200$ ⁠ mol, and $c = \dfrac{0.200}{0.1} = 2.00\ \mathrm{mol/L}$ ⁠. It is also written 2.00 M, “two molar”.

Hint
Moles first, $n = \dfrac{m}{M}$ ⁠; then $c = \dfrac{n}{V}$ ⁠, with $V$ ⁠ in litres.

Worked solution
1. Moles: $n = \dfrac{m}{M} = \dfrac{11.7}{58.44} = 0.200\ \mathrm{mol}$ ⁠.
2. Volume in litres: $V = 0.1\ \mathrm{L}$ ⁠.
3. Concentration: $c = \dfrac{n}{V} = \dfrac{0.200}{0.1} = 2.00$ ⁠.
4. The answer: 2.00 mol/L.
If you got 0.2, you forgot to divide by the volume, or divided by 1 L. If you got 117, you divided grams by the volume without turning them into moles.
Step 3 of 5
Open this step
Make a solution
Practice · Reach the target
Question
Change Salt added until the solution’s concentration is 0.50 mol/L, then press Check.

Goal
- 0.5 mol/L ± 0.01 mol/L
  At the start of the step: 1.711 mol/L
Setup, choices and feedback
The Scene at this step
Salt added
10 g
Water temperature
293 Kfixed
Stir
offfixed
Beaker / Particles
Beaker
Read from the Scene at
Dissolved solute: its moles over the 0.1 L (10⁻⁴ m³) of water/nodes/beaker/outputs/concentration
One setting that reaches it
Salt added
2.9 g
Feedback for a right answer
About 2.9 g: $0.5 \times 0.1 = 0.05$ ⁠ mol, and $0.05 \times 58.44 \approx 2.92$ ⁠ grams.

Feedback for a wrong answer
Not yet. $m = cVM$ ⁠.

Hint
How many moles do you need in 0.1 L, and how many grams is that?

Worked solution
1. Moles needed: $n = cV = 0.5 \times 0.1 = 0.05\ \mathrm{mol}$ ⁠.
2. Grams: $m = nM = 0.05 \times 58.44 = 2.92\ \mathrm{g}$ ⁠.
3. Set Salt added to 2.9 g.
Step 4 of 5
Open this step
In the lab
Practice · Calculate
Question
How many grams of $\mathrm{NaCl}$ ⁠ (58.44 g/mol) do you need to make 250 mL of a 0.40 mol/L solution?

Correct answer
5.84 g ± 0.05 g
Setup, choices and feedback
The Scene at this step
Salt added
10 gdefault
Water temperature
293 Kdefault
Stir
offdefault
Beaker / Particles
Beakerdefault
Accepted within
± 0.05 g
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 23.4 g ± 0.1 g
  Misconception: Treats a concentration as an amount, forgetting to multiply by the volume.
  Feedback from the catalogue
  A concentration is an amount per litre, not an amount. To know how much solute the solution holds, multiply by the volume: $n = cV$ ⁠. 250 mL of a 0.4 mol/L solution holds $0.4 \times 0.25 = 0.1$ ⁠ mol, not 0.4.
Feedback for a right answer
$n = 0.40 \times 0.250 = 0.100$ ⁠ mol, and $m = 0.100 \times 58.44 = 5.84$ ⁠ grams. In the lab you weigh out the salt, dissolve it, and top up with water to the 250 mL mark on a volumetric flask.

Hint
$n = cV$ ⁠, with $V$ ⁠ in litres.

Worked solution
1. Moles: $n = cV = 0.40 \times 0.250 = 0.100\ \mathrm{mol}$ ⁠.
2. Grams: $m = nM = 0.100 \times 58.44 = 5.84$ ⁠.
3. The answer: 5.84 g.
If you got 23.4 g, you treated the concentration as an amount: that is the salt for a whole litre, not for 250 mL.
Step 5 of 5
Open this step
Only what dissolved
Practice · Calculate
Question
The glass has 45 g of salt in 100 mL of water at 20 °C, and some of it is on the bottom. What is the solution’s concentration?

Correct answer
6.229 mol/L ± 2%
Setup, choices and feedback
The Scene at this step
Salt added
45 gfixed
Water temperature
293 Kfixed
Stir
offdefault
Beaker / Particles
Beaker
Accepted within
± 2%
Computed from the Scene at this step’s setting
Dissolved solute: its moles over the 0.1 L (10⁻⁴ m³) of water/nodes/beaker/outputs/concentration
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 7.7 mol/L ± 2%
  Misconception: Thinks any amount of solid will dissolve if you wait or stir long enough.
  Feedback from the catalogue
  There is a limit to how much the water can hold, the solubility: at 20 °C, 100 g of water holds at most about 36 g of salt. Beyond that the solution is saturated, and every extra gram stays on the bottom, however long you wait or stir.
Feedback for a right answer
Only 36.4 g dissolved, that is $\dfrac{36.4}{58.44} = 0.623$ ⁠ mol, and $c \approx 6.23\ \mathrm{mol/L}$ ⁠. That is the highest concentration a salt solution can reach at this temperature.

Hint
Only what has dissolved is in the solution. How much dissolved at 20 °C?

Worked solution
1. The solubility at 20 °C is 36.4 g per 100 g of water, so 36.4 g of the 45 g dissolved.
2. Moles dissolved: $n = \dfrac{36.4}{58.44} = 0.623$ ⁠.
3. Concentration: $c = \dfrac{0.623}{0.1} = 6.23$ ⁠.
4. The answer: 6.23 mol/L.
If you got 7.7, you counted the salt on the bottom too. It is not in the solution.

Lesson 4 of 4 · About 25 minutes

Dilution and separation

What changes when you add water to a solution and what stays the same, and how to take a mixture apart again: filtering, evaporation, crystallisation and distillation.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
A sample
Practice · Choose
Question
A glass holds 100 mL of salt solution at 2.0 mol/L. Half of it is poured into another glass. What is the concentration in the new glass?

Correct answer
- Correct 2.0 mol/L
Setup, choices and feedback
The Scene at this step
Salt added
11.7 g
Water temperature
293 K
Stir
offdefault
Beaker / Particles
Beaker
Choices
Shown in a random order.
- Correct 2.0 mol/L
- Wrong 1.0 mol/L
  Misconception: Thinks pouring out part of a solution lowers the concentration of what is poured or what is left.
  Feedback from the catalogue
  A solution is uniform: every drop of it has the same concentration. Pour off half and each half holds half the solute in half the volume, so the ratio, the concentration, does not change. Concentration changes only when solvent or solute is added or removed.
- Wrong 4.0 mol/L
  Feedback
  You neither added salt nor took out water: the ratio between them could not have grown.
Feedback for a right answer
The same concentration: half the volume holds half the salt, $\dfrac{0.1\ \mathrm{mol}}{0.05\ \mathrm{L}} = 2.0\ \mathrm{mol/L}$ ⁠. That is why you can take a small sample of a solution and measure the concentration of the whole.

Worked solution
1. In the original glass: $n = 2.0 \times 0.1 = 0.2$ ⁠ mol.
2. In the half poured off: $0.1$ ⁠ mol in 0.05 L.
3. $c = \dfrac{0.1}{0.05} = 2.0$ ⁠.
4. The answer: 2.0 mol/L.
Step 2 of 6
Open this step
Dilution
Practice · Calculate
Question
You take 10 mL of a 2.0 mol/L solution and add water up to 100 mL. What is the new concentration?

Correct answer
0.2 mol/L ± 0.005 mol/L
Setup, choices and feedback
The Scene at this step
Salt added
10 gdefault
Water temperature
293 Kdefault
Stir
offdefault
Beaker / Particles
Beakerdefault
Accepted within
± 0.005 mol/L
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 20 mol/L ± 0.05 mol/L
  Misconception: Turns the volume ratio upside down when diluting.
  Feedback from the catalogue
  When diluting, the moles of solute do not change, so $c_1 V_1 = c_2 V_2$ ⁠. Adding water makes the volume bigger, so the concentration gets smaller: $c_2 = c_1 \dfrac{V_1}{V_2}$ ⁠. If the concentration came out bigger after adding water, the ratio is upside down.
Feedback for a right answer
$c_2 = \dfrac{2.0 \times 10}{100} = 0.20\ \mathrm{mol/L}$ ⁠: ten times the volume, a tenth of the concentration. That is how a lab makes dilute solutions from a concentrated stock solution.

Hint
$c_1 V_1 = c_2 V_2$ ⁠.

Worked solution
1. Diluting does not change the moles: $c_1 V_1 = c_2 V_2$ ⁠.
2. The givens: $c_1 = 2.0$ ⁠, $V_1 = 10\ \mathrm{mL}$ ⁠, $V_2 = 100\ \mathrm{mL}$ ⁠.
3. $c_2 = \dfrac{2.0 \times 10}{100} = 0.20$ ⁠.
4. The answer: 0.20 mol/L.
The volumes can stay in millilitres, since their ratio does not depend on the unit. If you got 20, you turned the ratio upside down.
Step 3 of 6
Open this step
How much water
Practice · Calculate
Question
You have 50 mL of a 1.0 mol/L solution and need a 0.25 mol/L one. To what volume must you dilute it?

Correct answer
200 mL ± 1 mL
Setup, choices and feedback
The Scene at this step
Salt added
10 gdefault
Water temperature
293 Kdefault
Stir
offdefault
Beaker / Particles
Beakerdefault
Accepted within
± 1 mL
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 12.5 mL ± 0.5 mL
  Misconception: Turns the volume ratio upside down when diluting.
  Feedback from the catalogue
  When diluting, the moles of solute do not change, so $c_1 V_1 = c_2 V_2$ ⁠. Adding water makes the volume bigger, so the concentration gets smaller: $c_2 = c_1 \dfrac{V_1}{V_2}$ ⁠. If the concentration came out bigger after adding water, the ratio is upside down.
Feedback for a right answer
$V_2 = \dfrac{1.0 \times 50}{0.25} = 200\ \mathrm{mL}$ ⁠: you add 150 mL of water to the original solution.

Hint
$V_2 = \dfrac{c_1 V_1}{c_2}$ ⁠.

Worked solution
1. $c_1 V_1 = c_2 V_2$ ⁠, so $V_2 = \dfrac{c_1 V_1}{c_2}$ ⁠.
2. $V_2 = \dfrac{1.0 \times 50}{0.25} = 200$ ⁠.
3. The answer: 200 mL, so you add 150 mL of water.
If you got 12.5, you turned the ratio upside down: diluting always makes the volume bigger.
Step 4 of 6
Open this step
Filtering
Practice · Calculate
Question
The glass holds 45 g of salt in 100 g of water at 20 °C. You filter the mixture, then evaporate all the water from the filtrate, the liquid that came through the filter. How many grams of salt are left in the dish?

Correct answer
36.4 g ± 0.2 g
Setup, choices and feedback
The Scene at this step
Salt added
45 gfixed
Water temperature
293 Kfixed
Stir
offdefault
Beaker / Particles
Beaker
Accepted within
± 0.2 g
Computed from the Scene at this step’s setting
Solute dissolved in the 100 g of water/nodes/beaker/outputs/dissolved
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0 g ± 0.2 g
  Misconception: Thinks a filter can hold back a dissolved solute.
  Feedback from the catalogue
  A filter stops only solid particles bigger than its holes, such as crystals or sand. A dissolved substance is broken into single ions or molecules, far smaller than the holes, and it passes through with the water into the filtrate. To get it out you have to evaporate the water.
- 45 g ± 0.2 g
  Misconception: Thinks any amount of solid will dissolve if you wait or stir long enough.
  Feedback from the catalogue
  There is a limit to how much the water can hold, the solubility: at 20 °C, 100 g of water holds at most about 36 g of salt. Beyond that the solution is saturated, and every extra gram stays on the bottom, however long you wait or stir.
Feedback for a right answer
36.4 g: all the salt that dissolved passed through the filter, and was left behind when the water evaporated. The other 8.6 g stayed on the filter paper as crystals.

Worked solution
1. At 20 °C, 36.4 g dissolved and 8.6 g stayed solid.
2. The filter stops the crystals, but the dissolved ions pass through with the water.
3. When the filtrate is evaporated, 36.4 g of salt is left.
If you answered 0, you took the filter to stop dissolved salt too.
Step 5 of 6
Open this step
Crystallisation
Practice · Calculate
Question
39 g of salt has all dissolved in 100 g of boiling water. The solution is cooled to 20 °C. How many grams of crystals come out?

Correct answer
2.6 g ± 0.1 g
Setup, choices and feedback
The Scene at this step
Salt added
39 gfixed
Water temperature
293 Kfixed
Stir
offdefault
Beaker / Particles
Beaker
Accepted within
± 0.1 g
Computed from the Scene at this step’s setting
Solute left as a pile on the bottom/nodes/beaker/outputs/undissolved
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 39 g ± 0.1 g
  Misconception: Thinks hotter water always dissolves much more of anything.
  Feedback from the catalogue
  It depends on the substance. The solubility of most solids rises with temperature, but table salt’s hardly does: 36 g at 20 °C and 39 g at 100 °C. And the solubility of gases falls when it is hot: that is why a warm fizzy drink goes flat fast.
Feedback for a right answer
Only 2.6 g: $39 - 36.4$ ⁠. This is crystallisation by cooling, and it works well for substances whose solubility drops a lot in the cold. For table salt it is better to evaporate the water, as in the evaporation ponds at the Dead Sea.

Worked solution
1. At 20 °C the water holds at most 36.4 g.
2. The solution has 39 g, so the excess comes out: $39 - 36.4 = 2.6$ ⁠.
3. The answer: 2.6 g.
If you answered 39 g, you assumed the solubility drops a lot when it cools. For table salt it hardly changes.
Step 6 of 6
Open this step
Distillation
Practice · Choose
Question
How do you get fresh water, with no salt, from seawater?

Correct answer
- Correct Distillation: boil it, and condense the steam in a separate vessel
Setup, choices and feedback
The Scene at this step
Salt added
11.7 g
Water temperature
373 K
Stir
offdefault
Beaker / Particles
Beaker
Choices
Shown in a random order.
- Correct Distillation: boil it, and condense the steam in a separate vessel
- Wrong Filtering through filter paper
  Misconception: Thinks a filter can hold back a dissolved solute.
  Feedback from the catalogue
  A filter stops only solid particles bigger than its holes, such as crystals or sand. A dissolved substance is broken into single ions or molecules, far smaller than the holes, and it passes through with the water into the filtrate. To get it out you have to evaporate the water.
- Wrong Evaporating it in an open dish
  Feedback
  Evaporating leaves the salt in the dish, and the water goes into the air. To keep the water you must cool the steam and collect it: that is distillation.
- Wrong Leave it to stand until the salt settles
  Misconception: Thinks any amount of solid will dissolve if you wait or stir long enough.
  Feedback
  Dissolved salt never settles: the ions are spread evenly, and the molecules’ motion keeps them that way.
Feedback for a right answer
Distillation uses the difference in boiling points: water boils at 100 °C and the salt stays solid. The steam passes through a condenser, condenses, and is collected as pure water. Israel’s desalination plants mostly use another method, reverse osmosis, which forces the water through a membrane the ions cannot pass.

Worked solution
1. The salt is dissolved, so a filter will not stop it and nothing will settle.
2. The difference between the water and the salt is in their boiling points.
3. Boil it: only the water turns to steam.
4. Condense the steam in another vessel: you get pure water, and the salt stays behind.

Across the unit · 11 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Dissolving

Where it went

Question

Correct answer

The Scene at this step

Choices

Worked solution

Inside the water

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Mass is conserved

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Worked solution

Solution or mixture

Question

Correct answer

The Scene at this step

Choices

Worked solution

Solubility

Saturated

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Worked solution

Stirring

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Hot water

Question

Correct answer

The Scene at this step

Once the prediction is locked in, the Scene changes

Worked solution

Heat until it all dissolves

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Gases

Question

Correct answer

The Scene at this step

Choices

Worked solution

Concentration

Grams per litre

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Moles per litre

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Make a solution

Question

Goal

The Scene at this step

One setting that reaches it

Hint