Solutions & Dissolved Particles — Problem set, Grade 11

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

Review the key ideas

Solutions & Solubility

A solution forms when a solute spreads evenly through a solvent; solubility is the maximum that will dissolve at a given temperature, and it shifts with temperature, pressure, and the polarity of the substances.

Concentration & Molarity

Molarity (M) measures concentration as moles of solute per liter of solution (M = mol ÷ L); to dilute a solution, use M₁V₁ = M₂V₂ because the moles of solute stay the same.

Electrolytes & Colligative Properties

Colligative changes depend approximately on the number of dissolved particles, not merely solute grams.

Problem 1

A student dissolves 8 g of sugar in 200 g of water. Identify the solute and the solvent.
Answer and explanation

Solute = sugar; solvent = water.

The solvent is the substance present in the larger amount that does the dissolving — the 200 g of water. The solute is the substance being dissolved, present in the smaller amount — the 8 g of sugar.

Problem 2

The solubility of potassium nitrate (KNO₃) at 40°C is 64 g per 100 g of water. How many grams will dissolve in 150 g of water at 40°C?
Answer and explanation

96 g

150 g of water is 1.5 times 100 g (150 ÷ 100 = 1.5). Multiply the solubility: 64 g × 1.5 = 96 g of KNO₃ will dissolve.

Problem 3

At 20°C the solubility of sodium chloride is 36 g per 100 g of water. If you stir 50 g of NaCl into 100 g of water at 20°C, how many grams remain undissolved at the bottom?
Answer and explanation

14 g

Only 36 g of NaCl can dissolve in 100 g of water at this temperature (the solution becomes saturated). So 50 g − 36 g = 14 g of NaCl remains undissolved.

Problem 4

What is the molarity of a solution containing 0.50 mol of solute dissolved in enough water to make 2.0 L of solution?
Answer and explanation

0.25 M

Molarity = moles of solute ÷ liters of solution = 0.50 mol ÷ 2.0 L = 0.25 M.

Problem 5

You dilute 25.0 mL of 4.0 M hydrochloric acid to a final volume of 100.0 mL. What is the new molarity?
Answer and explanation

1.0 M

Use the dilution equation M₁V₁ = M₂V₂, solving for M₂ = M₁V₁ ÷ V₂ = (4.0 M × 25.0 mL) ÷ 100.0 mL = 100 ÷ 100 = 1.0 M.

Problem 6

A solution is made by dissolving 20.0 g of sodium hydroxide (NaOH, molar mass 40.0 g/mol) in enough water to make 0.250 L of solution. What is its molarity?
Answer and explanation

2.00 M

First convert grams to moles: moles = mass ÷ molar mass = 20.0 g ÷ 40.0 g/mol = 0.500 mol. Then molarity = moles ÷ liters = 0.500 mol ÷ 0.250 L = 2.00 M.

Problem 7

How much water must you add to 50.0 mL of 6.0 M hydrochloric acid to dilute it to 2.0 M? (Find the final volume first.)
Answer and explanation

Final volume = 150 mL, so you must add 100 mL of water.

Use M₁V₁ = M₂V₂ to find the final volume: V₂ = M₁V₁ ÷ M₂ = (6.0 M × 50.0 mL) ÷ 2.0 M = 300 ÷ 2.0 = 150 mL. Since you started with 50.0 mL, the water added is 150 mL − 50 mL = 100 mL.

Problem 8

Would stirring necessarily increase grams dissolved at saturation and fixed T?
Answer and explanation

It can speed reaching equilibrium but does not generally change the equilibrium solubility.

Response criteria

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

Problem 9

Chapter review: 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

Find M for.20 mol dissolved to a final.50L.
Answer and explanation

M=n/V=.40 mol/L. Dilution lowers M without changing solute amount if no reaction or loss occurs.

Response criteria

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

Problem 11

Chapter review: 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 12

For i=2, Kf=1.86, b=.10 find freezing depression.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

0.372 °C

2×1.86×.10.

Problem 13

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 14

Chapter review: 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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