Exercises on Solutions (with commented template)

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Test your knowledge of chemical solutions with the 10 questions Next. Check the comments after the feedback to clear your doubts about the topic.

question 1

A solution can be defined as

a) Pure substance at a given temperature and pressure.
b) Heterogeneous mixture with uniform properties in all phases.
c) Mixture of at least two substances with uniform appearance.
d) Dispersion of a solid material in a liquid.

Correct alternative: c) Mixture of at least two substances with uniform appearance.

A solution can be defined as a system formed by a homogeneous mixture of two or more substances. Therefore, the components of a uniform mixture are not differentiated with the naked eye or with the use of an optical microscope.

Examples of solutions are:

  • Mixture of water and acetic acid (vinegar);
  • Mixture of water and salt;
  • Mixture of water and sugar.

question 2

In the solutions:

I. Water and salt
II. water and sugar
III. Sodium bicarbonate and water

The substances salt, sugar and bicarbonate are classified as

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a) solvent
b) solute
c) Colloid
d) Dispersant

Correct alternative: b) Solute.

Solute is a component that is dissolved in the substance in greater quantity, which is called a solvent.

In the solutions presented, water is the dispersant and the other components are the dispersed ones.

In homogeneous mixtures, the average size of the dispersed particles does not exceed 1 nanometer. Therefore, the components of the solutions are not perceptible to the naked eye and not even with the use of an optical microscope.

question 3

Look at the mixtures below.

I. atmospheric air
II. Ethyl alcohol 96º GL
III. gelatin in water
IV. Granite
v. Blood
SAW. Milk of magnesia

Which of the systems are NOT classified as solutions?

a) I, II and VI
b) II, III and IV
c) III, IV, V and VI
d) All except I.

Correct alternative: c) III, IV, V and VI.

I. Atmospheric air is a homogeneous mixture of gases.
II. Ethyl alcohol is a homogeneous mixture of water and alcohol.
III. Gelatin in water is a colloidal dispersion.
IV. Granite is a heterogeneous mixture of stones.
v. Blood is a heterogeneous mixture. The components can be seen using the microscope.
SAW. Milk of magnesia is an aqueous suspension.

know more about homogeneous and heterogeneous mixtures.

question 4

500 ml of water were added to a solution with a volume of 500 ml and 5 g of sodium chloride. Regarding the final solution, analyze the following statements.

I. The final solution is a dilution.
II. The final volume of the solution is 1L.
III. The common concentration of the final solution is 5 g/L.
SAW. The number of moles of solute was halved in the final solution.

The statements are correct:

a) Only II
b) I and II
c) I, II and III
d) All are correct

Correct alternative: c) I, II and III.

I. CORRECT. Dilution consists of adding pure solvent to the preexisting solution.
II. CORRECT. In a dilution, the final volume is calculated by the formula Vf = Vi + VThe

Vf = Vi + VThe
Vf = 0.5 L + 0.5 L
Vf = 1L

III. CORRECT. After a dilution, the final concentration of the solution is determined by the formula Ci.Vi = Cf.Vf

The common concentration of the starting solution is:

Çi = mass (g)/volume of solution (L)
Çi = 5 g/0.5L
Çi = 10 g/L

Therefore, the common concentration of the final solution is:

Çi.Vi = Cf.Vf
10 g/L. 0.5 L = Cf. 1L
5g/1 L = Cf
Çf = 5 g/L

IV. WRONG. In a dilution, the number of moles of solute remains constant.

know more about dilution.

question 5

A solution was prepared by dissolving a salt that has a solubility of 120 g/L in water at 25°C as follows form: 140 g of the solute were added to one liter of water, whose temperature was 35 ºC, and the mixture was cooled to 25 ºC The solution obtained can be classified as:

a) saturated
b) unsaturated
c) supersaturated
d) concentrated

Correct alternative: c) supersaturated.

The solubility coefficient indicates the maximum capacity of the solute that dissolves in a given amount of solvent. Therefore, 120 g of the salt presented in the statement forms a saturated solution with one liter of water at 25 °C.

However, the dissolution capacity can be altered by temperature. As the solvent was heated, the increase in temperature increased its dissolution capacity. Thus, when returning to the temperature of 25 °C, we have a supersaturated solution, in which the amount of solute is greater than the solubility coefficient.

know more about solubility.

question 6

When evaporating the solvent from 500 mL of a solution with a common concentration of 5 g/L, what is the mass of solute obtained?

a) 0.5 g
b) 1 g
c) 2.5 g
d) 5 g

Correct alternative: c) 2.5 g.

The common concentration, also called the concentration in g/L, is the ratio of the mass of solute in a volume of solution.

Mathematically, the common concentration is expressed through the formula: C = m/V


C: common concentration;
m: mass of the solute;
V: volume of solution.

As the common concentration is given in g/L, in this case we need to convert the unit of volume before determining the mass of the solute.

As 1 L contains 1000 mL, so 500 mL corresponds to 0.5 L.

table row with cell with C space equals 5 space g slash L end of cell blank blank blank blank row with cell with m space equals space? end of cell blank cell with C equals end of cell cell with m in bottom frame end of cell double right arrow row with cell with V space equals space 0 comma 5 space L end of cell blank blank V blank row with blank blank blank blank blank end of table table row with blank blank blank blank blank row with cell with m equal to end of cell cell with C space. space V end of cell equals cell with 5 space g slash up risk L space. space 0 comma 5 space diagonal up risk L end of cell cell equal to space 2 comma 5 g end of cell row with blank blank blank blank blank row with blank blank blank blank blank end of table

Thus, when evaporating the solvent from the solution with a concentration of 5 g/L, 2.5 g of solute was obtained.

know more about common concentration.

question 7

What is the resulting molarity of 250 mL of a solution prepared by dissolving 0.395 g of potassium permanganate (KMnO4), whose molar mass is 158 g/mol?

a) 0.01 M
b) 0.02 M
c) 0.03 M
d) 0.04 M

Correct alternative: a) 0.01 M

The molarity formula is M = n1/V


no1 = number of moles of solute (in mole);
V = volume of solution (in L).

Knowing that the formula for potassium permanganate is KMnO4 and its molar mass is 158 g/mol, the first step is to calculate the number of moles of 0.395 g of KMnO4. For this, we can apply the rule of three.

1 mole - 158 g
x moles - 0.395 g
x = 0.0025 moles

Now, we calculate the molarity of the solution.

M = n1/V
M = 0.0025 mol/0.25 L
M = 0.01 M

know more about molarity.

question 8

What is the resulting molality of the solution prepared with 2 L of water, of density 1 g/mL, in which 80 g of hydrochloric acid (HCl) was dissolved, whose molar mass is 36.5 g/mol?

a) 0.4 mol/kg
b) 1.1 mol/kg
c) 2.4 mol/kg
d) 1.5 mol/kg

Correct alternative: b) 1.1 mol/kg.

Molality (W) or molal concentration is the result of the amount of solute matter per mass of solvent.

W = n1/m2


W = molality (given in mol/kg)
no1 = amount of substance of the solute (given in mol)
m2 = mass of solvent (given in kg)

The first step in solving the question is to calculate the number of moles of the solute:

no1 = m1/M1
no1 = 80 g/36.5 g/mol
no1 = 2.2 moles

Now we calculate the value of the solvent mass (m2) from the density formula:

d = m/v → m = d. v → m2 = (1.0 g/ml). (2000 mL) → m2 = 2000 g or 2.0 kg of water

Applying the values ​​found in the density formula, we have:

W = n1/m2
W = 2.2 mol/2.0 kg
W = 1.1 mol/kg or 1.1 mol

know more about molality.

question 9

(UFRS) The solubility of caustic soda (NaOH) in water, as a function of temperature, is given in the table below.

Temperature (º C) 20 30 40 50
Solubility (grams/100 g of H2O 109 119 129 145

Considering solutions of NaOH in 100 g of water, it is correct to state that:

a) at 20 °C, a solution with 120 g of NaOH is concentrated.
b) at 20 °C, a solution with 80 g of NaOH is diluted.
c) at 30 °C, a solution with 11.9 g of NaOH is concentrated.
d) at 30 °C, a solution with 119 g of NaOH is supersaturated.
e) at 40 °C, a solution with 129 g of NaOH is saturated.

Correct alternative: e) at 40 °C, a solution with 129 g of NaOH is saturated.

a) WRONG. At 20 °C, a solution with 120 g of NaOH is saturated with a bottom body, since the maximum dissolved solute at this temperature is 109.

b) WRONG. At 20 °C, a solution with 80 g of NaOH is unsaturated because the amount of solute is less than the solubility coefficient.

c) WRONG. The amount of solute is less than the maximum dissolution capacity at the observed temperature.

d) WRONG. The solution with 119 g of NaOH at 30 °C is saturated.

e) CORRECT. The solution has the maximum amount of solute fully dissolved by the solvent.

question 10

(Mackenzie) A typical example of a supersaturated solution is:

a) natural mineral water.
b) homemade whey.
c) refrigerant in a closed container.
d) 46°GL alcohol.
e) vinegar.

Correct alternative: c) refrigerant in a closed container.

a) WRONG. Mineral water is a solution, that is, a homogeneous mixture with dissolved salts.

b) WRONG. Homemade whey is a solution of water, sugar and salt in defined amounts.

c) CORRECT. Soda is a mixture of water, sugar, concentrates, coloring, aroma, preservatives and gas. The carbon dioxide (CO2) dissolved in the refrigerant is forming a supersaturated solution.

Increasing the pressure increases the solubility of the gas, causing much more gas to be added to the refrigerant than performing the same operation at atmospheric pressure.

One of the characteristics of supersaturated solutions is that they are unstable. We can see that when opening the bottle with soda, a small part of the gas escapes, because the pressure inside the container has decreased.

d) WRONG. Alcohol 46 °GL is a hydrated alcohol, that is, it contains water in its composition.

e) WRONG. Vinegar is a solution of acetic acid (C2H5OH) and water.

Gain more knowledge with the contents:

  • chemical solutions
  • solute and solvent
  • Exercises on common concentration

Bibliographic references

BROWN, Theodore; LEMAY, H. Eugene; BURSTEN, Bruce E. Chemistry: the core science. 9 ed. Prentice Hall, 2005.

FELTRE, Ricardo. Fundamentals of Chemistry: vol. single. 4th ed. Sao Paulo: Moderna, 2005.

PERUZZO. F.M.; CORNER. E.L., Chemistry in everyday life, volume 1, 4th edition, modern ed, São Paulo, 2006.

  • chemical solutions
  • Solute and Solvent: what they are, differences and examples
  • Solubility
  • Exercises on Common Concentration with commented feedback
  • Molarity or Molar Concentration: what it is, formula and how to calculate
  • Solution concentration
  • Common concentration: what it is, how to calculate it and solved exercises
  • Dilution of solutions

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