Common concentration: what it is, how to calculate and solved exercises

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

Mathematically, the common concentration is expressed through the formula:

C = m/v

Where,

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

The most commonly used unit for common concentration is g/L, as the mass of solute is calculated in grams (g) and the volume of solution in liters (L).

For example, sea water has a large amount of dissolved salts and sodium chloride (NaCl), popularly known as table salt, is one of them.

The concentration of sodium chloride in seawater is, on average, 2.5 g/L. So, there are 2.5 grams of salt, which is the solute, in every 1 liter of seawater, which is the solution.

How to calculate common concentration?

A solution is a homogeneous mixture formed by a substance in smaller amount, called solute, dissolved in a solvent, which is in larger amount.

Hence, common concentration refers to the amount of solute in a given volume of solution. The more solute dissolved in the solution, the more concentrated it is. Otherwise, that is, low solute indicates that the solution is diluted.

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Example 1: What is the concentration in g/L of a solution of silver iodide (AgI) containing 2.6 g in 1L of solution?

When given the amount of solute and the volume of the solution, we simply plug the values into the common concentration formula to find its value.

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Therefore, in a 2.6 g/L solution of silver iodide there are 2.6 g of solute in every 1 liter of solution.

Example 2: When evaporating the solvent from 500 mL of saline solution with a concentration of 6 g/L, what is the mass of solute obtained?

Note that in some calculations, we can find the common concentration described so that we can calculate the mass of solute.

It is also necessary to pay attention to the units. As the common concentration is given in g/L, in this case we need to convert the unit of volume before applying the formula.

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

table row with cell with C space equals 6 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 6 space g slash up risk L space. space 0 comma 5 space diagonal up risk L end of cell cell equal to space 12 g end from cell line with blank blank blank blank blank line with blank blank blank blank blank end of table

Thus, when evaporating the solvent from the solution with a concentration of 6 g/L, 12 g of solute was obtained.

Gain more knowledge with the contents:

Solution concentration
molar concentration
Dilution of solutions

Solved common concentration exercises

Use the following questions to check the knowledge acquired previously.

question 1

(Unicamp) The solvent is completely evaporated from 250 mL of an aqueous solution of MgCl₂ 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

see answer

Correct alternative: d) 2.0.

table row with cell with C space equal to 8 comma 0 space g slash L end of cell blank blank blank blank row with cell with m space equal to 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 250 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 8 space g slash up risk L space. space 0 comma 25 space diagonal up risk L end of cell cell equal to space 2 g end from cell line with blank blank blank blank blank line with blank blank blank blank blank end of table

question 2

See the image below.

In this scheme, container 3 contains a

a) heterogeneous mixture
b) pure substance
c) Solution
d) colloid mixture

see answer

Correct alternative: c) Solution.

Container 3 contains a solution, which is a homogeneous mixture formed by solute and solvent, where salt is the solute and water is the solvent.

Generally, we can say that the solute is the component in the smallest amount and the solvent is present in the largest amount.

Container 1 presents a heterogeneous mixture with 3 phases and container 2 corresponds to a heterogeneous mixture with 2 phases.

question 3

A solution was prepared by dissolving 4.0 g of magnesium chloride MgCl₂until reaching a concentration of 2.0 g/L. What volume of solution was prepared?

a) 1 L
b) 2 L
c) 4 L
d) 6 L

see answer

Correct alternative: b) 2 L.