Common Concentration: exercises with commented feedback

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Common concentration is the amount of solute, in grams, in 1 liter of solution.

Mathematically, the common concentration is expressed by: straight C space equal to numerator mass space space solute over denominator volume space space solution end of fraction

1. (Mackenzie) What is the concentration, in g/L, of the solution obtained by dissolving 4 g of sodium chloride in 50 cm3 of water?

a) 200 g/L
b) 20 g/L
c) 0.08 g/L
d) 12.5 g/L
e) 80 g/L

Correct alternative: e) 80 g/L.

1st step: Transform the volume unit of cm3 to L.

Knowing that 1 cm3 = 1 mL, then we have:

table row with cell with 1000 mL space end of cell minus cell with 1 straight space L end of cell row with cell with 50 mL space end of cell minus straight V row with blank blank blank row with straight V equal to cell with numerator 50 horizontal space line stroke mL space. space 1 straight space L over denominator 1000 horizontal space risk mL end fraction end of cell line with straight V equals cell with 0 comma 05 straight space L end of cell end of table

Step 2: Apply the data in the common concentration formula:

straight C space equal to straight m with left parenthesis straight g right parenthesis subscript end of subscript over straight V with left parenthesis straight L right parenthesis subscript end of straight subscript C space equal to numerator 4 straight space g over denominator 0 comma 05 straight space L end of fraction straight C space equal to 80 straight space g divided by straight L

2. (Mackenzie) There are five containers containing aqueous solutions of sodium chloride.

solutions

It is correct to say that:

a) container 5 contains the least concentrated solution.
b) container 1 contains the most concentrated solution.
c) only containers 3 and 4 contain solutions of equal concentration.
d) the five solutions have the same concentration.
e) container 5 contains the most concentrated solution.

Correct alternative: d) the five solutions have the same concentration.

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Applying the Common Concentration Formula straight C space equal to straight m with left parenthesis straight g right parenthesis subscript end of subscript over straight V with left parenthesis straight L right parenthesis subscript end of subscript for each of the containers, we have:

1 2 3 4 5
straight C space equal to numerator 0 comma 5 straight space g over denominator 2 straight space L end of fraction straight C space equal to 0 comma 25 straight space g divided by straight L straight C space equal to numerator 0 comma 75 straight space g over denominator 3 straight space L end of fraction straight C space equal to 0 comma 25 straight space g divided by straight L straight C space equal to numerator 1 comma 25 straight space g over denominator 5 straight space L end of fraction straight C space equal to 0 comma 25 straight space g divided by straight L straight C space equal to numerator 2 comma 0 straight space g over denominator 8 straight space L end of fraction straight C space equal to 0 comma 25 straight space g divided by straight L straight C space equal to numerator 2 comma 5 straight space g over denominator 10 straight space L end of fraction straight C space equal to 0 comma 25 straight space g divided by straight L

From the calculations performed, we see that all solutions have the same concentration.

3. (UFPI) The new traffic legislation provides for a maximum limit of 6 decigrams of alcohol, C2H5OH, per liter of blood from the driver (0.6 g/L). Considering that the average percentage of alcohol ingested that remains in the blood is 15% by mass, identify, for an adult with an average weight of 70 kg whose blood volume is 5 liters, the maximum number of beer cans (volume = 350 mL) ingested without the established limit being outdated. Additional information: the beer has 5% alcohol by volume, and the alcohol density is 0.80 g/mL.

to 1
b) 2
c) 3
d) 4
e) 5

Correct alternative: a) 1.

Question data:

  • Maximum allowed blood alcohol limit: 0.6 g/L
  • Percentage of ingested alcohol remaining in the blood: 15%
  • Blood volume: 5 L
  • Beer can volume: 350 mL
  • Percentage of alcohol in beer: 5%
  • Alcohol density: 0.80 g/mL

1st step: Calculate the mass of alcohol in 5 L of blood.

straight C space equal to straight m with left parenthesis straight g right parenthesis subscript end of subscript over straight V with left parenthesis straight L right parenthesis subscript end of subscript space double arrow to the right straight m with left parenthesis straight g right parenthesis subscript end of subscript space equal to straight C space. straight space V with left parenthesis straight L right parenthesis subscript end of subscript
straight m with left parenthesis straight g right parenthesis subscript end of subscript space equal to straight C space. straight space V with left parenthesis straight L right parenthesis subscript end of straight subscript m with left parenthesis straight g right parenthesis subscript end of subscript space equal to 0 comma 6 straight space g divided by straight L space. space 5 straight space L straight m with left parenthesis straight g right parenthesis subscript end of subscript space equal to 3 comma 0 straight space g space of space alcohol

2nd step: Calculate the total alcohol mass, as only 15% was absorbed into the bloodstream.

table row with cell with straight m with 1 subscript end of cell minus cell with 100 percent sign end of cell row with cell with 3 comma 0 straight space g end of cell minus cell with 15 percent sign end of cell row with blank blank blank row with cell with straight m with 1 subscript end of cell equals cell with numerator 3 comma 0 space straight g space. space 100 percent sign over denominator 15 percent sign end of fraction space end of cell row with blank blank blank row with cell with straight m with 1 subscript end of cell equals cell with 20 straight space g space alcohol end of cell end of table

3rd step: Calculate the volume of alcohol present in the beer.

straight d space equal to straight m over straight V space double right arrow space V space equal to straight m over straight d
straight V space equal to numerator 20 straight space g over denominator 0 comma 8 straight space g divided by mL end of fraction straight V space equal to 25 space mL

Step 4: Calculate the maximum volume of beer that can be consumed.

table row with cell with 25 mL space end of cell minus cell with 5 percent sign end of cell row with cell with straight V with 2 subscript end of cell minus cell with 100 percent sign end of cell row with blank blank blank row with cell with straight V with 2 subscript end of cell equal to cell with numerator 25 mL space space. space 5 percent sign over denominator 100 percent sign end of fraction space end of cell row with blank blank blank row with cell with straight V with 2 subscript end of cell equals cell with 500 mL space space beer end of cell end of table

5th step: Interpretation of results.

The maximum volume of beer that a person can drink so that the concentration of alcohol in the blood does not exceed 0.6 g/L is 500 ml.

Each beer contains 350 mL and when consuming two cans, the volume is 700 mL, which exceeds the established volume. As such, the most a person can ingest is one can.


4. (UNEB) The homemade serum consists of an aqueous solution of sodium chloride (3.5 g/L) and sucrose (11 g/L). The masses of sodium chloride and sucrose needed to prepare 500 ml of homemade serum are, respectively:

a) 17.5 g and 55 g
b) 175 g and 550 g
c) 1750mg and 5500mg
d) 17.5 mg and 55 mg
e) 175 mg and 550 mg

Correct alternative: c) 1750mg and 5500mg.

Calculate the mass of sodium chloride

1st step: Transform the volume unit from mL to L.

table row with cell with 1000 mL space end of cell minus cell with 1 straight space L end of cell row with cell with 500 mL space end of cell minus straight V row with blank blank blank row with straight V equal to cell with numerator 500 horizontal space line stroke mL space. space 1 straight space L over denominator 1000 horizontal space risk mL end fraction end of cell line with straight V equals cell with 0 comma 5 straight space L end of cell end of table

2nd step: Calculate the mass in grams.

straight C space equal to straight m with left parenthesis straight g right parenthesis subscript end of subscript over straight V with left parenthesis straight L right parenthesis subscript end of subscript space double arrow to the right straight m with left parenthesis straight g right parenthesis subscript end of subscript space equal to straight C space. straight space V with left parenthesis straight L right parenthesis subscript end of subscript
straight m with left parenthesis straight g right parenthesis subscript end of subscript space equal to straight C space. straight space V with left parenthesis straight L right parenthesis subscript end of straight subscript m with left parenthesis straight g right parenthesis subscript end of subscript space equal to 3 comma 5 straight space g divided by straight L space. space 0 comma 5 straight space L straight m with left parenthesis straight g right parenthesis subscript end of subscript space equal to 1 comma 75 straight space g space NaCl space

3rd step: Transform the found value to milligrams.

table row with cell with 1 straight space g end of cell minus cell with 1000 mg space end of cell row with cell with 1 comma 75 straight space g end of cell minus cell with straight m with NaCl subscript end of cell row with blank blank blank row with cell with straight m with NaCl subscript end of cell equal to cell with numerator 1 comma 75 straight space g space. space 1000 space mg over denominator 1 straight space g end of fraction space end of cell row with blank blank blank row with cell with straight m with NaCl subscript end of cell equals cell with 1750 mg space space alcohol end of cell end of table

Calculate the mass of sucrose

1st step: Calculate the mass in grams.

Knowing that 500 mL = 0.5 L, then we have:

straight C space equal to straight m with left parenthesis straight g right parenthesis subscript end of subscript over straight V with left parenthesis straight L right parenthesis subscript end of subscript space double arrow to the right straight m with left parenthesis straight g right parenthesis subscript end of subscript space equal to straight C space. straight space V with left parenthesis straight L right parenthesis subscript end of subscript
straight m with left parenthesis straight g right parenthesis subscript end of subscript space equal to straight C space. straight space V with left parenthesis straight L right parenthesis subscript end of straight subscript m with parenthesis left straight g right parenthesis subscript end of subscript space equal to 11 straight space g divided by straight L space. space 0 comma 5 straight space L straight m with left parenthesis straight g right parenthesis subscript end of subscript space 5 comma 5 straight space g sucrose space space

2nd step: Transform the found value to milligrams.

table row with cell with 1 straight space g end of cell minus cell with 1000 mg space end of cell row with cell with 5 comma 5 straight space g end of cell minus cell with straight m with sucrose subscript end of cell row with blank blank blank row with cell with straight m with sucrose subscript end of cell equal to cell with numerator 5 comma 5 straight space g space. space 1000 space mg over denominator 1 straight space g end of fraction space end of cell row with blank blank blank row with cell with straight m with sucrose subscript end of cell equals cell with 5500 mg space space sucrose end of cell end of table
5. (PUC-Campinas) Totally evaporate the solvent from 250 ml of an aqueous MgCl solution2 of concentration 8.0 g/L. How many grams of solute are obtained?

a) 8.0
b) 6.0
c) 4.0
d) 2.0
e) 1.0

Correct alternative: d) 2.0.

1st step: Transform the volume unit from mL to L.

table row with cell with 1000 mL space end of cell minus cell with 1 straight space L end of cell row with cell with 250 mL space end of cell minus straight V row with blank blank blank row with straight V equal to cell with numerator 250 horizontal space line stroke mL space. space 1 straight space L over denominator 1000 horizontal space risk mL end fraction end of cell line with straight V equals cell with 0 comma 250 straight space L end of cell end of table

2nd step: Calculate the mass of magnesium chloride (MgCl2).

straight C space equal to straight m with left parenthesis straight g right parenthesis subscript end of subscript over straight V with left parenthesis straight L right parenthesis subscript end of subscript space double arrow to the right straight m with left parenthesis straight g right parenthesis subscript end of subscript space equal to straight C space. straight space V with left parenthesis straight L right parenthesis subscript end of subscript
straight m with left parenthesis straight g right parenthesis subscript end of subscript space equal to straight C space. straight space V with left parenthesis straight L right parenthesis subscript end of straight subscript m with left parenthesis straight g right parenthesis subscript end of subscript space equal to 8 comma 0 straight space g divided by straight L space. space 0 comma 25 straight space L straight m with left parenthesis straight g right parenthesis subscript end of subscript space equal to 2 straight space g space MgCl space with 2 subscript
6. (Mackenzie) The mass of the four main salts that are dissolved in 1 liter of sea water is equal to 30 g. In a marine aquarium, containing 2.106 cm3 of this water, the amount of salts dissolved in it is:

a) 6.0. 101 kg
b) 6.0. 104 kg
c) 1.8. 102 kg
d) 2.4. 108 kg
e) 8.0. 106 kg

Correct alternative: a) 6.0. 101 kg.

1st step: Calculate the mass of dissolved salts in the aquarium.

Knowing that 1 L = 1000 mL = 1000 cm3, we have:

table row with cell with 1000 cm space cubed end of cell minus cell with 30 straight space g end of cell row with cell with 2.10 to the power of 6 cm space cubed end of cell minus straight m row with blank blank blank row with straight m equal to cell with numerator 2.10 to the power of 6 horizontal strikeout space over cm to cube end of strikeout space. space 30 straight space g over denominator 1000 horizontal strikeout space over cm to cube end of strikeout end of fraction end of cell row with straight V equals cell with 60 space 000 straight space g end of cell end of table

Step 2: Transform mass unit from grams to kilograms.

table row with cell with 1 space kg end of cell minus cell with 1000 straight space g end of cell row with cell with straight m with 2 subscript end of cell minus cell with 60000 straight space g end of cell row with blank blank blank row with cell with straight m with 2 subscript end of cell equal to cell with numerator 60000 straight space g space. space 1 space kg over denominator 1000 straight space g end of fraction end of cell row with cell with straight m with 2 subscript end of cell equals cell with 60 kg space end of cell end of table

3rd step: Transform the result to scientific notation.

As a number in scientific notation it has the format N. 10no, to turn 60 kg into scientific notation we "walk" with the comma and place it between 6 and 0.

We have N = 6.0 and since we are only walking one decimal place, the value of n is 1 and the correct answer is: 6.0. 101 kg.

7. (UFPI) An analgesic in drops must be administered in amounts of 3 mg per kilogram of body mass, however, it cannot exceed 200 mg per dose. Knowing that each drop contains 5 mg of analgesic, how many drops should be given to a 70 kg patient?

Correct answer: 40 drops.

Question data:

  • Recommended analgesic dose: 3 mg/kg
  • Amount of analgesic in drop: 5 mg of analgesic
  • patient weight: 70 kg

1st step: Calculate the amount of analgesic according to the patient's weight.

table row with cell with 3 space mg end of cell minus cell with 1 space kg end of cell row with straight m minus cell with 70 kg space end of cell row with blank blank blank row with straight m equal to cell with numerator 3 mg space space. space 70 space kg over denominator 1 space kg end of fraction end of cell row with straight V equals cell with 210 space mg end of cell end of table

The calculated amount exceeds the maximum dose. Therefore, 200 mg must be administered, which corresponds to the permitted limit.

2nd step: Calculate the amount of analgesic drop.

table row with cell with 5 space mg end of cell minus cell with 1 space drop end of cell row with cell with 200 space mg end of cell minus straight V row with blank blank blank row with straight V equal to cell with numerator 200 mg space space. space 1 space drop over denominator 5 space mg end of fraction end of cell line with straight V equals cell with 40 space drops end of cell end of table

8. (Enem) A certain station treats about 30 000 liters of water per second. To avoid the risk of fluorosis, the maximum concentration of fluorides in this water should not exceed about 1.5 milligrams per liter of water. The maximum amount of this chemical species that can be safely used, in the volume of water treated in one hour, at this station, is:

a) 1.5 kg
b) 4.5 kg
c) 96 kg
d) 124 kg
e) 162 kg

Correct alternative: e) 162 kg.

Question data:

  • Treated water: 30,000 L/s
  • Fluoride concentration: 1.5 mg/L

1st step: Transform hour into minutes.

table row with Hour blank Minutes blank Seconds row with cell with 1 straight line h end of cell cell with right arrow with straight x space 60 superscript end of cell cell with 60 min space end of cell cell with right arrow with straight x space 60 superscript end of cell cell with 3600 straight space s end of cell end of table

2nd step: Calculate fluoride mass in 30000 L/s.

table row with cell with 1 comma 5 space mg end of cell minus cell with 1 straight space L end of cell row with cell with straight space m with 1 subscript end of cell minus cell with 30000 straight space L end of cell row with blank blank blank row with cell with straight m with 1 subscript end of cell equal to cell with numerator 1 comma 5 space mg space. space 30000 horizontal space straight line L over denominator 1 horizontal space straight line L end of fraction end of cell row with cell with straight m with 1 subscript end of cell equals cell with 45000 space mg end of cell end of table

3rd step: Calculate the mass for the time of 1 h (3600 s).

table row with cell with 45000 space mg end of cell minus cell with 1 straight space s end of cell row with cell with straight space m with 2 subscript end of cell minus cell with 3600 straight space s end of cell row with blank blank blank row with cell with straight space m with 2 subscript end of cell equal to cell with numerator 45000 space mg space. space 3600 horizontal space straight line s over denominator 1 horizontal space line line s end of fraction end of cell row with cell with straight space m with 2 subscript end of cell equals cell with 162000000 space mg end of cell end of table

4th step: Transform the unit of mass from mg to kg.

table row with milligram blank Gram blank Kilo row with cell with 162000000 space mg end of cell cell with right arrow divided by space 1000 superscript end of cell cell with 162000 straight space g end of cell cell with right arrow with divided by space 1000 superscript end of cell cell with 162 kg space end of cell end of table

9. (UFRN) One of the economic potentials of Rio Grande do Norte is the production of sea salt. Sodium chloride is obtained from sea water in salt pans built near the coast. In general, sea water travels through several crystallization tanks up to a determined concentration. Suppose that, in one of the steps of the process, a technician took 3 samples of 500 mL from a crystallization, carried out evaporation with each sample and noted the resulting salt mass in table a follow:

Sample Sample volume (mL) Salt mass (g)
1 500 22
2 500 20
3 500 24

The average concentration of the samples will be:

a) 48 g/L
b) 44 g/L
c) 42 g/L
d) 40 g/L

Correct alternative: b) 44 g/L.

1st step: Transform the volume unit from mL to L.

table row with cell with 1000 mL space end of cell minus cell with 1 straight space L end of cell row with cell with 500 mL space end of cell minus straight V row with blank blank blank row with straight V equal to cell with numerator 500 horizontal space line stroke mL space. space 1 straight space L over denominator 1000 horizontal space risk mL end fraction end of cell line with straight V equals cell with 0 comma 5 straight space L end of cell end of table

Step 2: Apply the common concentration formula straight C space equal to straight m with left parenthesis straight g right parenthesis subscript end of subscript over straight V with left parenthesis straight L right parenthesis subscript end of subscript for each of the samples.

1 2 3
straight C with 1 subscript space equal to numerator 22 straight space g over denominator 0 comma 5 straight space L end of fraction straight C with 1 subscript space equal to 44 straight space g divided by straight L straight C with 2 subscript space equal to numerator 20 straight space g over denominator 0 comma 5 straight space L end of fraction straight C with 2 subscript space equal to 40 straight space g divided by straight L straight C with 3 subscript space equal to numerator 24 straight space g over denominator 0 comma 5 straight space L end of fraction straight C with 3 subscript space equal to 48 straight space g divided by straight L

3rd step: Calculate the average concentration.

straight C with straight m subscript space equal to numerator space straight C with 1 subscript space plus straight space C with 2 subscript space plus straight space C with 3 subscript over denominator 3 end of fraction straight C with straight m subscript space equal to numerator space 44 straight space g divided by straight L plus space 40 straight space g divided by straight L space plus space 48 straight space g divided by straight L over denominator 3 end of fraction straight C with straight m subscript space equal to 44 straight space g divided by straight L

10. (Fuvest) Consider two cans of the same soda, one in the “diet” version and the other in the common version. Both contain the same volume of liquid (300 mL) and have the same mass when empty. The composition of the refrigerant is the same in both, except for one difference: the common version contains a certain amount of sugar, while the “diet” version does not contain sugar (only negligible mass of a sweetener artificial). By weighing two closed soda cans, the following results were obtained:

Sample Mass (g)
Can with regular soda 331.2 g
Can with "diet" soda 316.2 g

Based on these data, it can be concluded that the concentration, in g/L, of sugar in regular soft drinks is approximately:

a) 0.020
b) 0.050
c) 1.1
d) 20
e) 50

Correct alternative: e) 50.

1st step: Calculate the sugar mass.

As the only difference between soft drinks is the mass of sugar, as it is only present in the common version, we can find it by subtracting the masses given from each sample.

straight M with sugar subscript space equal to straight M with straight R common space subscript end of subscript space minus straight space M with straight R space diet subscript end of subscript straight space M with subscript sugar space equal to 331 comma 2 straight space g space minus space 316 comma 2 straight space g straight space M with subscript sugar space equal to 15 straight space g

2nd step: Transform the volume unit from mL to L.

table row with cell with 1000 mL space end of cell minus cell with 1 straight space L end of cell row with cell with 300 mL space end of cell minus straight V row with blank blank blank row with straight V equal to cell with numerator 300 horizontal space streak mL space. space 1 straight space L over denominator 1000 horizontal space risk mL end fraction end of cell line with straight V equals cell with 0 comma 3 straight space L end of cell end of table

3rd step: Calculate the sugar concentration.

straight C space equal to straight m with left parenthesis straight g right parenthesis subscript end of subscript over straight V with left parenthesis straight L right parenthesis subscript end of subscript straight space C space equal to numerator 15 straight space g over denominator 0 comma 3 straight space L end of fraction straight space C space equal to 50 straight space g divided by straight L

To gain more knowledge about chemical solutions, see also these texts.:

  • Solute and solvent
  • Dilution of solutions
  • Molarity
  • Molality
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