Solutions & Dissolved Particles — Chapter Practice, Grade 11

Study saturation, concentration, conductivity, and particle-count effects.

Problem 1

In a solution of sugar dissolved in water, which part is the solvent?
  1. the sugar
  2. the water
  3. the sweetness
  4. the glass
Answer and explanation

the water

The solvent does the dissolving and is present in the larger amount — here, the water.

Problem 2

A solution that contains the maximum amount of dissolved solute at a given temperature is described as…
  1. nonpolar
  2. unsaturated
  3. saturated
  4. dilute
Answer and explanation

saturated

A saturated solution holds all the solute it can at that temperature.

Problem 3

The solubility of potassium nitrate at 40°C is 64 g per 100 g of water. How many grams will dissolve in 300 g of water at 40°C?
Answer and explanation

192 g

300 g of water is 3 times 100 g, so 64 g × 3 = 192 g dissolves.

Problem 4

Stirring a solute and grinding it into a fine powder both make it dissolve faster, yet neither changes the maximum amount that can dissolve. Explain why.
Answer and explanation

Stirring and grinding increase the surface area where solvent meets solute, so dissolving happens at a faster rate. But the maximum that can dissolve — the solubility — depends on the temperature and the identities of the substances, not on how quickly you get there.

Problem 5

At 20°C the solubility of a certain salt is 34 g per 100 g of water. If you stir 50 g of the salt into 100 g of water at 20°C, how many grams remain undissolved at the bottom?
Answer and explanation

16 g

Only 34 g can dissolve in 100 g of water, so 50 g − 34 g = 16 g stays undissolved.

Problem 6

At solubility 20 g/100 g water, how much can 250 g water dissolve under the model? Show the reasoning, not just the conclusion.
Hints

Solubility is temperature-specific.

Answer and explanation

20×250/100=50 g.

Problem 7

Why can warming increase some solid solubilities but reduce dissolved gas amounts?
Hints

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

Answer and explanation

Different dissolution thermodynamics apply; there is no universal temperature direction for all substances.

Problem 8

Changing the rate need not change the equilibrium capacityChanging the rate need not change the equilibrium capacityDissolved solute mass (g)051015012345Time (relative)Unstirred modelStirred model
Original illustrative approach-to-equilibrium records; the curves are not experimental fits and apply to the stated fixed-condition model.
Interpret the supplied visual. What is the main difference between the two selected curves, and what stays the same?
Answer and explanation

The stirred model approaches the common 15 g plateau faster. The selected final equilibrium capacity is unchanged.

Dissolution rate is how quickly solute enters solution, while equilibrium solubility specifies a final capacity under stated conditions. Stirring and smaller particles can speed approach to saturation by improving transport or exposed area without changing the final equilibrium amount in the selected model. Temperature can affect both rate and equilibrium, with the direction depending on the system. A faster curve reaching the same plateau supports a rate difference, not a claim that stirring created extra solvent capacity.

The stirred model approaches the common 15 g plateau faster. The selected final equilibrium capacity is unchanged.

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 9

At a supplied temperature, solubility is 20.0 g solute per 100 g water. Add 30.0 g to 100 g water and allow equilibrium with excess solid. Find dissolved and undissolved masses and explain what stirring changes.
Answer and explanation

Under the stated model, 20.0 g dissolves and 10.0 g remains solid. Stirring can speed equilibration but does not change this selected equilibrium capacity at fixed conditions.

Under the stated model, 20.0 g dissolves and 10.0 g remains solid. Stirring can speed equilibration but does not change this selected equilibrium capacity at fixed conditions.

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.

Problem 10

Molarity is defined as…
  1. grams of solute per liter of solution
  2. liters of solution per mole of solute
  3. moles of solute per gram of solvent
  4. moles of solute per liter of solution
Answer and explanation

moles of solute per liter of solution

Molarity (M) equals moles of solute divided by liters of solution.

Problem 11

What is the molarity of a solution containing 3.0 mol of potassium chloride in 1.5 L of solution?
Answer and explanation

2 M

M = 3.0 mol ÷ 1.5 L = 2.0 M.

Problem 12

Describe how you would prepare 1.0 L of a 0.50 M NaCl solution. (The molar mass of NaCl is 58.5 g/mol.)
Answer and explanation

You need 0.50 mol/L × 1.0 L = 0.50 mol of NaCl, which is 0.50 mol × 58.5 g/mol ≈ 29.3 g. Weigh out about 29.3 g of NaCl, dissolve it in some water in a 1.0 L volumetric flask, then add water up to the 1.0 L line and mix.

Problem 13

You dilute 100.0 mL of a 2.0 M solution with water to a final volume of 500.0 mL. What is the new molarity?
Answer and explanation

0.4 M

M₂ = M₁V₁ ÷ V₂ = (2.0 × 100.0) ÷ 500.0 = 0.40 M.

Problem 14

What is the molarity of a solution made by dissolving 117 g of NaCl (molar mass 58.5 g/mol) in enough water to make 2.00 L of solution?
Answer and explanation

1 M

Moles = 117 ÷ 58.5 = 2.00 mol; M = 2.00 mol ÷ 2.00 L = 1.00 M.

Problem 15

Find molarity of 0.030 mol in 0.150 L solution. Show the reasoning, not just the conclusion.
Hints

Final solution volume is not necessarily the volume of added water.

Answer and explanation

M=0.030/0.150=0.20 mol/L.

Problem 16

A solution has 2 g solute in 100 g total solution. Find mass percent.
Hints

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

Answer and explanation

2/100×100%=2%; use solution mass, not solvent mass.

Problem 17

Dilution conserves solute amount only under the stated no-reaction conditionsDilution conserves solute amount only under the stated no-reaction conditionsBefore dilutionAfter dilutionc₁V₁ = nsolute = c₂V₂Same solute amount; larger solution volume lowers concentration.
Original equal-count dilution model; dot spacing and solvent volume are schematic rather than a molecular-scale simulation.
Interpret the supplied visual. Why does doubling final volume halve concentration in this selected dilution?
Answer and explanation

The unchanged solute amount is distributed through twice the solution volume, so n/V halves.

Dilution adds solvent while retaining the same amount of the specified solute, assuming no reaction, precipitation or loss. From n=cV, this gives c₁V₁=c₂V₂. The final volume V₂ is the total diluted solution volume, not merely the added solvent volume. The same expression should not be used automatically for reacting acid and base solutions, because the named solute amount can change. Draw a before-and-after entity inventory to identify what is actually conserved.

The unchanged solute amount is distributed through twice the solution volume, so n/V halves.

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 18

Prepare 250. mL of 0.200 M solution by diluting a 1.00 M stock without reaction or loss. Find stock volume and explain whether 250. mL is added water volume.
Answer and explanation

V₁=c₂V₂/c₁=(0.200)(250.)/1.00=50.0 mL stock. Bring the mixture to 250. mL final solution volume; do not interpret that as adding exactly 250. mL water.

V₁=c₂V₂/c₁=(0.200)(250.)/1.00=50.0 mL stock. Bring the mixture to 250. mL final solution volume; do not interpret that as adding exactly 250. mL water.

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.

Problem 19

Find molality of .30 mol solute in .60 kg solvent.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

0.5 mol/kg

.30/.60=.50.

Problem 20

Why can dissolved sugar fail to conduct well?
Answer and explanation

Neutral sugar molecules do not supply many mobile ions.

Response criteria

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

Problem 21

Find ideal freezing depression for i=1, Kf=1.86, b=.30 mol/kg.
Hints

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

Answer and explanation

0.558 °C

1×1.86×.30=.558.

Problem 22

Compare molarity and molality denominators.
Answer and explanation

Molarity uses solution volume; molality uses solvent mass in kilograms.

Response criteria

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

Problem 23

Find ΔT_f for m=0.20, i=1 and K_f=1.86°C·kg/mol. Show the reasoning, not just the conclusion.
Hints

Use molality, not molarity, for the standard freezing-point relation.

Answer and explanation

ΔT_f=iK_fm=0.372°C.

Problem 24

Why can an ionic solute conduct electricity while glucose solution usually conducts poorly?
Hints

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

Answer and explanation

Mobile ions carry charge; dissolved neutral glucose does not supply comparable ionic carriers.

Problem 25

A phase-equilibrium explanation is stronger than saying particles block freezingA phase-equilibrium explanation is stronger than saying particles block freezingDepression magnitude ΔTf (K)00.511.522.500.20.40.6Molality b (mol/kg solvent)Selected dilute model: slope = iKf = 3.72 K·kg/mol.
Original calculated ideal-model curve, not observed freezing data or a complete water–salt phase diagram.
Interpret the supplied visual. What does the slope describe in the selected linear model, and why should it not be extrapolated without limit?
Answer and explanation

It is iKf. At larger concentration, nonideality and other phase behavior can invalidate the simple dilute relation.

A dissolved solute changes the solvent’s chemical potential and therefore the conditions for phase equilibrium. In the selected dilute model, more effective solute particles give a larger freezing-temperature shift. This is not a literal wall of solute particles physically blocking every water molecule from crystallizing. Real phase diagrams, eutectic limits and concentration changes during freezing add detail beyond the straight-line approximation. Interpret the graph within its stated range and avoid extending it indefinitely to impossible temperatures.

It is iKf. At larger concentration, nonideality and other phase behavior can invalidate the simple dilute relation.

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 26

Using Kf=1.86 K·kg/mol, compare ideal freezing depressions for 0.100 mol glucose and 0.100 mol CaCl₂, each in 0.500 kg water. Assume i=1 and 3 respectively.
Answer and explanation

Both formula molalities are 0.200 mol/kg. Glucose gives ΔTf=0.372 K and ideal CaCl₂ gives 1.116 K≈1.12 K. Actual effective i and nonideality can change the measured comparison.

Both formula molalities are 0.200 mol/kg. Glucose gives ΔTf=0.372 K and ideal CaCl₂ gives 1.116 K≈1.12 K. Actual effective i and nonideality can change the measured comparison.

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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