PERCENTAGE SOLUTIONS: WHAT THEY CONSIST OF, TYPES AND EXAMPLES - CHEMISTRY - 2024

The percentage solutions are those whose solute concentration is expressed by 100 mL of solution. For example, a percentage solution of 5 g / 100 ml is equivalent to the expression 5% (w / v). Thus, their concentrations are expressed with the use of percentages.

There are many ways to express the concentration of a solute in a solution. Among them are molarity, which indicates the concentration of the solute as the number of moles per liter of solution; molality, moles of solute per kilogram of solvent; and normality, solute equivalents per liter of solution.

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The concentration of a solute can also be expressed as a percentage. It is the simplest way of expressing the concentration, since it does not require calculations or knowledge of certain parameters such as the molecular weight of the solute, its equivalent weight or the characteristics of its ionic dissociation.

Alcoholic beverages are examples of percentage solutions. On their labels they specify the degree of alcohol, which is not more than its concentration expressed per 100 mL of the liquid in the bottle. The higher the alcoholic strength, the more intense its effects on the body.

What are the percentage solutions?

Percent solutions or solutions indicate the amount of a solute dissolved per one hundred parts of solution. This type of expressing the concentration of solutions is often used in commercial products to indicate their chemical composition. However, it is of little use in teaching and research laboratories.

Types of percentage solutions

Percentage Mass - Volume

Indicate the mass of solute dissolved in 100 cm ³ of Solution. The mathematical expression for the calculation of this concentration is:

% m / v = (mass of solute in g / volume of the solution in cm ³) x 100

Mass Percentage - Mass

Indicate the mass of solute contained in 100 g of solution. Mass is a property that does not vary with temperature or pressure, so this way of expressing concentration is the preferred one in chemical analysis reports. The mathematical expression for its calculation is:

% m / m = (mass of solute in g / mass of solution in g) x 100

Percentage Volume - Volume

Indicates the volume of a liquid dissolved in 100 ml of solution. The liquids must be miscible and it must be possible to correct the volume changes that can occur when the liquids are mixed. The mathematical expression for its calculation is:

% v / v = (volume of solute in cm ³ / volume of solution in cm ³) x 100

Examples

Example 1

A potassium chloride (KCl) solution is prepared by mixing 10 g of the compound with 120 g of water. Express the concentration of the solution in% m / m.

The mass of the solute corresponds to 10g of KCl, and that of the solvent, 120g of water. Adding both masses the solution is obtained: 130g. Therefore, it is enough to apply the mathematical equation:

% KCl m / m = (10 g KCl / 130 g solution) x 100

7.69

Example 2

50 cm ³ of 100% acetic acid (CH ₃ COOH) are mixed with 130 cm ³ of water. What will be the percentage of acetic acid v / v

In the same way as the previous example, the volume of the solution must be obtained by adding the volumes of the solute and the solvent. Thus, the Vsol is (50 + 130) cm ³, and the% v / v therefore:

% Acetic acid v / v = (50 cm ³ /180 cm ³) x 100

27.77% v / v

Example 3

It is desired to prepare 1L of a solution of glycerin at 8% v / v in alcohol. Find the volume of the glycerin and the alcohol in the solution.

Using the percentage you can determine how much glycerin is dissolved in the 1L bottle:

= Volume of glycerol (8 cm ³ /100 cm ³) x 1000 cm ³

80 cm ³

Remember that 1L of solution is equal to 1000cm ³ or 1000mL. Once the 80 cm ³ of glycerin has been obtained, these must be subtracted from the total volume of solution to know how much alcohol was used in the preparation:

Volume of alcohol = 1000 cm ³ - 80 cm ³

Example 4

8 g of paraffin is dissolved in 50 ml of glycerol, which has a density of 1.26 g / cm ³. Find the concentration of the paraffin in glycerol in% m / m.

Density data allows you to determine how much a liquid weighs. With this value the mass of glycerol used in this percentage solution can be calculated:

Glycerol mass = glycerol volume x glycerol density

Glycerol mass = 50 cm ³ x 1.26 g / cm ³

63 g

To calculate the% m / m it is only necessary to divide the mass of paraffin by the total mass of solution:

Mass of paraffin in 100 g of glycerol = (8 g / 63g) x 100 g

12.70

That is, for every 100g of solution there are 12.70g of paraffin.

Example 5

A concentrated solution of NaOH at 28% m / m has a density of 1.15 g / cm3. Find the grams of NaOH present in one liter of the reagent.

To take advantage of% m / m, you need to express the solution by mass. Once again, its density allows to calculate it:

Reagent mass = volume x density

1000 cm ³ x 1.15 g / cm ³

1150 g

Therefore, the mass of NaOH in the solution is:

Mass of NaOH in 1000 cm ³ = (28 g / 100g) x 1150 g

322 g

Example 6

400 mL of 10% m / v NaCl are mixed with 600 mL of 5% m / v NaCl. Find the concentration of the resulting solution, expressing it in% m / v.

Two solutions are mixed to make a third. All three have the NaCl solute in common. Therefore, you must calculate the mass of the salt that were taken from the first two solutions:

Mass of NaCl solution 1 = volume1 x concentration 1

400 cm ³ x (10 g / 100 cm ³)

40 g

Mass of NaCl solution 2 = volume2 x concentration2

600 cm ³ x (5 g / 100 cm ³)

30 g

And the total volume when mixing the two aliquots (assuming they are additive) is:

Volume of the mixture: 400 mL + 600 mL

1000 mL

That is, in 1L of solution there are 70g of NaCl (30 + 40). So to calculate the% m / m percentage, instead of 1L per 100mL of solution, you must express:

Mass of NaCl in 100 cm ³ = (70 g / 1000 cm ³) x 100 cm ³

7 g

NaCl concentration in the mixture = 7% m / m

References

1. Miranda M. Molina B. (March 9, 2013). What are percentage solutions? Taken from: quimicacb20gpo457sasi.blogspot.com
2. Cristina Andrade Guevara. (2012). Percentage solutions.. Taken from: roa.uveg.edu.mx
3. Prof. N. De Leon. (sf). Concentration: Volume / Volume. Taken from: iun.edu
4. Aus e Tute. (sf). Weight / Volume Percentage Concentration. Taken from: ausetute.com.au
5. Flores, J. Chemistry. (2002) Editorial Santillana.