Solutions & Dissolved Particles — Chapter quiz · Physical Science: Chemistry, Grade 11

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

Problem 1

In a solution of salt dissolved in water, which part is the solvent?
  1. the beaker
  2. the salt
  3. the water
  4. the dissolved ions
Answer and explanation

the water

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

Problem 2

The solubility of a salt is 36 g per 100 g of water at a given temperature. How many grams will dissolve in 250 g of water at that temperature?
Answer and explanation

90 g

250 g of water is 2.5 times 100 g, so 36 g × 2.5 = 90 g dissolves.

Problem 3

What is the molarity of a solution containing 3.0 mol of solute dissolved in enough water to make 1.5 L of solution?
Answer and explanation

2 M

M = moles ÷ liters = 3.0 mol ÷ 1.5 L = 2.0 M.

Problem 4

You dilute 50.0 mL of 4.0 M hydrochloric acid with water to a final volume of 200.0 mL. What is the new molarity?
Answer and explanation

1 M

M₂ = M₁V₁ ÷ V₂ = (4.0 × 50.0) ÷ 200.0 = 200 ÷ 200 = 1.0 M.

Problem 5

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 1.00 L of solution?
Answer and explanation

2 M

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

Problem 6

Explain the following claim and supply a valid example or counterexample: Dissolving faster is not the same as having a larger equilibrium solubility.
Answer and explanation

It can speed reaching equilibrium but does not generally change the equilibrium solubility. Dissolving faster is not the same as having a larger 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 7

Independent chapter check: 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 8

Explain the following claim and supply a valid example or counterexample: Molarity uses final solution volume, not the volume of water added.
Answer and explanation

M=n/V=.40 mol/L. Dilution lowers M without changing solute amount if no reaction or loss occurs. Molarity uses final solution volume, not the volume of water added.

Response criteria

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

Problem 9

Independent chapter check: 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 10

For i=1,Kb=.512,b=.50 find boiling elevation.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

0.256 °C

.512×.50.

Problem 11

Why might a measured salt solution differ from i=2 exactly?
Answer and explanation

Interactions between ions and nonideal activity affect the effective particle behavior.

Response criteria

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

Problem 12

Independent chapter check: 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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