The concentration of solutions. Methods for expressing the concentration of solutions

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1 millimol per liter [mmol/L] = 0.001 mol per liter [mol/L]

Initial value

Converted value

moles per meter³ moles per liter moles per centimeter³ moles per millimeters decimeter molar millimolar micromolar nanomolar picomolar femtomolar attomolar zeptomolar yoctomolar

More about molar concentration

General information

The concentration of a solution can be measured in many ways, such as the ratio of the mass of the solute to the total volume of the solution. In this article, we will look at molar concentration, which is measured as the ratio between the amount of substance in moles to the total volume of the solution. In our case, a substance is a soluble substance, and we measure the volume for the entire solution, even if other substances are dissolved in it. Amount of substance is the number of elementary constituents, such as atoms or molecules of a substance. Since even a small amount of a substance usually contains a large number of elementary components, special units, moles, are used to measure the amount of a substance. One mole is equal to the number of atoms in 12 g of carbon-12, that is, it is approximately 6 × 10²³ atoms.

It is convenient to use moths if we are working with an amount of a substance so small that its amount can be easily measured with home or industrial devices. Otherwise, one would have to work with very large numbers, which is inconvenient, or with very small weights or volumes, which are difficult to find without specialized laboratory equipment. Atoms are most often used when working with moles, although other particles, such as molecules or electrons, can also be used. It should be remembered that if not atoms are used, then this must be indicated. Sometimes molar concentration is also called molarity.

Molarity should not be confused with molality. Unlike molarity, molality is the ratio of the amount of solute to the mass of the solvent, and not to the mass of the entire solution. When the solvent is water, and the amount of solute is small compared to the amount of water, then molarity and molality are similar in meaning, but otherwise they usually differ.

Factors affecting molar concentration

The molar concentration depends on temperature, although this dependence is stronger for some and weaker for other solutions, depending on what substances are dissolved in them. Some solvents expand with increasing temperature. In this case, if the substances dissolved in these solvents do not expand with the solvent, then the molar concentration of the entire solution decreases. On the other hand, in some cases, with increasing temperature, the solvent evaporates, and the amount of the solute does not change - in this case, the concentration of the solution will increase. Sometimes the opposite happens. Sometimes a change in temperature affects how a solute dissolves. For example, some or all of the solute ceases to dissolve and the concentration of the solution decreases.

Units

Molar concentration is measured in moles per unit volume, such as moles per liter or moles per cubic meter. Moles per cubic meter is an SI unit. Molarity can also be measured using other units of volume.

How to find molar concentration

To find the molar concentration, you need to know the amount and volume of a substance. The amount of a substance can be calculated using the chemical formula of that substance and information about the total mass of that substance in solution. That is, to find out the amount of the solution in moles, we find out from the periodic table the atomic mass of each atom in the solution, and then we divide the total mass of the substance by the total atomic mass of the atoms in the molecule. Before adding together the atomic mass, make sure that we multiply the mass of each atom by the number of atoms in the molecule we are considering.

You can also do the calculations in reverse order. If the molar concentration of the solution and the formula of the solute are known, then you can find out the amount of solvent in the solution, in moles and grams.

Examples

Find the molarity of a solution of 20 liters of water and 3 tablespoons of soda. In one tablespoon - about 17 grams, and in three - 51 grams. Baking soda is sodium bicarbonate whose formula is NaHCO₃. In this example, we'll use atoms to calculate molarity, so we'll find the atomic masses of the sodium (Na), hydrogen (H), carbon (C), and oxygen (O) constituents.

Na: 22.989769
H: 1.00794
C: 12.0107
O:15.9994

Since the oxygen in the formula is O₃, it is necessary to multiply the atomic mass of oxygen by 3. We get 47.9982. Now add the masses of all atoms and get 84.006609. The atomic mass is indicated in the periodic table in atomic mass units, or a. e. m. Our calculations are also in these units. One a. e.m. is equal to the mass of one mole of a substance in grams. That is, in our example, the mass of one mole of NaHCO₃ is 84.006609 grams. In our task - 51 grams of soda. We find the molar mass by dividing 51 grams by the mass of one mole, that is, by 84 grams, and we get 0.6 moles.

It turns out that our solution is 0.6 moles of soda dissolved in 20 liters of water. We divide this amount of soda by the total volume of the solution, that is, 0.6 mol / 20 l \u003d 0.03 mol / l. Since a large amount of solvent and a small amount of solute were used in the solution, its concentration is low.

Let's consider another example. Find the molar concentration of one sugar cube in a cup of tea. Table sugar is made up of sucrose. First, let's find the weight of one mole of sucrose, the formula of which is C₁₂H₂₂O₁₁. Using the periodic table, we find the atomic masses and determine the mass of one mole of sucrose: 12 × 12 + 22 × 1 + 11 × 16 = 342 grams. There are 4 grams of sugar in one cube of sugar, which gives us 4/342 = 0.01 moles. There are about 237 milliliters of tea in one cup, so the concentration of sugar in one cup of tea is 0.01 moles / 237 milliliters × 1000 (to convert milliliters to liters) = 0.049 moles per liter.

Application

Molar concentration is widely used in calculations related to chemical reactions. The branch of chemistry that calculates the ratios between substances in chemical reactions and often works with moles is called stoichiometry. The molar concentration can be found from the chemical formula of the final product, which then becomes a soluble substance, as in the soda solution example, but you can also first find this substance from the formulas of the chemical reaction during which it is formed. To do this, you need to know the formulas of the substances involved in this chemical reaction. Having solved the chemical reaction equation, we find out the formula of the molecule of the solute, and then we find the mass of the molecule and the molar concentration using the periodic table, as in the examples above. Of course, it is possible to perform calculations in reverse order, using information about the molar concentration of a substance.

Let's consider a simple example. This time we mix baking soda with vinegar to see an interesting chemical reaction. Both vinegar and baking soda are easy to find - you probably have them in your kitchen. As mentioned above, the formula for baking soda is NaHCO₃. Vinegar is not a pure substance, but a 5% solution of acetic acid in water. The formula for acetic acid is CH₃COOH. The concentration of acetic acid in vinegar can be more or less than 5%, depending on the manufacturer and the country in which it is made, as the concentration of vinegar varies from country to country. In this experiment, you do not have to worry about the chemical reactions of water with other substances, since water does not react with soda. We only care about the volume of water when we later calculate the concentration of the solution.

First, we solve the equation for the chemical reaction between soda and acetic acid:

NaHCO₃ + CH₃COOH → NaC₂H₃O₂ + H₂CO₃

The reaction product is H₂CO₃, a substance that, due to low stability, enters into a chemical reaction again.

H₂CO₃ → H₂O + CO₂

As a result of the reaction, we get water (H₂O), carbon dioxide (CO₂) and sodium acetate (NaC₂H₃O₂). We mix the resulting sodium acetate with water and find the molar concentration of this solution, just as before we found the concentration of sugar in tea and the concentration of soda in water. When calculating the volume of water, it is necessary to take into account the water in which acetic acid is dissolved. Sodium acetate is an interesting substance. It is used in chemical heating pads, such as hand warmers.

Using stoichiometry to calculate the amount of substances that enter into a chemical reaction, or reaction products for which we will later find the molar concentration, it should be noted that only a limited amount of a substance can react with other substances. This also affects the amount of the final product. If the molar concentration is known, then, on the contrary, it is possible to determine the amount of starting products by the reverse calculation method. This method is often used in practice, in calculations related to chemical reactions.

When using recipes, whether in cooking, in making medicines, or in creating the ideal environment for aquarium fish, it is necessary to know the concentration. In everyday life, it is most often convenient to use grams, but in pharmaceuticals and chemistry, molar concentration is more often used.

In pharmaceuticals

When creating drugs, the molar concentration is very important, since it determines how the drug affects the body. If the concentration is too high, then the drugs can even be fatal. On the other hand, if the concentration is too low, then the drug is ineffective. In addition, concentration is important in the exchange of fluids across cell membranes in the body. When determining the concentration of a liquid that must either pass or, conversely, not pass through the membranes, either the molar concentration is used, or it is used to find osmotic concentration. Osmotic concentration is used more often than molar concentration. If the concentration of a substance, such as a drug, is higher on one side of the membrane than on the other side of the membrane, such as inside the eye, then the more concentrated solution will move across the membrane to where the concentration is lower. This flow of solution across the membrane is often problematic. For example, if fluid moves into the interior of a cell, for example, into a blood cell, then it is possible that due to this fluid overflow, the membrane will be damaged and rupture. Leakage of fluid from the cell is also problematic, as this will disrupt the performance of the cell. Any drug-induced flow of fluid through the membrane out of or into the cell is desirable to prevent, and to do this the concentration of the drug is sought to be similar to that of a fluid in the body, such as blood.

It is worth noting that in some cases the molar and osmotic concentrations are equal, but this is not always the case. It depends on whether the substance dissolved in water has broken down into ions in the process electrolytic dissociation. The osmotic concentration calculation takes into account particles in general, while the molar concentration calculation takes into account only certain particles, such as molecules. Therefore, if, for example, we are working with molecules, but the substance has decomposed into ions, then the molecules will be less than the total number of particles (including both molecules and ions), and hence the molar concentration will be lower than the osmotic one. To convert the molar concentration to osmotic concentration, you need to know the physical properties of the solution.

In the manufacture of medicines, pharmacists also take into account tonicity solution. Tonicity is a property of a solution that depends on concentration. Unlike osmotic concentration, tonicity is the concentration of substances that the membrane does not let through. The process of osmosis causes solutions of higher concentration to move into solutions of lower concentration, but if the membrane prevents this movement by not allowing the solution to pass through, then there is pressure on the membrane. Such pressure is usually problematic. If a drug is intended to enter the blood or other body fluid, then the tonicity of the drug must be balanced against the tonicity of the body fluid to avoid osmotic pressure on the membranes in the body.

To balance tonicity, drugs are often dissolved in isotonic solution. An isotonic solution is a solution of table salt (NaCL) in water at a concentration that balances the tonicity of the fluid in the body and the tonicity of the mixture of this solution and the drug. Usually isotonic solution is stored in sterile containers and infused intravenously. Sometimes it is used in its pure form, and sometimes - as a mixture with medicine.

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It can be expressed both in dimensionless units (fractions, percentages) and in dimensional quantities (mass fractions, molarity, titers, mole fractions).

Concentration is the quantitative composition of the solute (in specific units) per unit volume or mass. The solute is named X, and the solvent - S. Most often I use the concept of molarity (molar concentration) and mole fraction.

1. (or percentage concentration of a substance) is the ratio of the mass of a solute m to the total mass of the solution. For a binary solution consisting of a solute and a solvent:

ω - mass fraction of the dissolved substance;

m in-va- mass of dissolved substance;

mr-ra is the mass of the solvent.

The mass fraction is expressed in fractions of a unit or as a percentage.

2. Molar concentration or molarity is the number of moles of a solute in one liter of solution V:

C- molar concentration of a dissolved substance, mol / l (it is also possible to designate M, for example, 0.2 MHCl);

V- the volume of the solution, l.

The solution is called molar or unimolar, if 1 mole of a substance is dissolved in 1 liter of solution, decimolar- 0.1 mol of substance is dissolved, centomolar- 0.01 mol of substance is dissolved, millimolar- 0.001 mol of a substance is dissolved.

3. Molar concentration(molality) of solution C(x) shows the number of moles n solute in 1 kg of solvent m:

C(x) - molality, mol/kg;

n- amount of dissolved substance, mol;

mr-la- mass of solvent, kg.

4. - substance content in grams in 1 ml of solution:

T- titer of dissolved substance, g/ml;

m in-va- mass of the dissolved substance, g;

V r-ra- volume of solution, ml.

5. - dimensionless quantity, equal to the ratio of the amount of solute n to the total amount of substances in solution:

N- molar fraction of the dissolved substance;

n- amount of dissolved substance, mol;

n r-la- the amount of solvent substance, mol.

The sum of the mole fractions must equal 1:

N(X) + N(S) = 1.

where N(X) X;

N(S) - mole fraction of solute S.

Sometimes, when solving problems, it is necessary to move from one unit of expression to another:

ω(X) - mass fraction of the dissolved substance, in%;

M(X) is the molar mass of the solute;

ρ = m/(1000 V) is the density of the solution.6. - the number of gram equivalents of a given substance in one liter of solution.

Gram equivalent of substance- the number of grams of a substance, numerically equal to its equivalent.

Equivalent- this is a conventional unit, equivalent to one hydrogen ion in acid-base reactions or one electron in redox reactions.

Abbreviations are used to record the concentration of such solutions. n or N. For example, a solution containing 0.1 mol-eq / l is called decinormal and is written as 0.1 n.

C N- normal concentration, mol-eq/l;

z- equivalence number;

V r-ra- the volume of the solution, l.

Solubility substances S - the maximum mass of a substance that can be dissolved in 100 g of a solvent:

Solubility factor- the ratio of the mass of a substance that forms a saturated solution at a specific temperature to the mass of the solvent:

One of the basic units in the International System of Units (SI) is the unit of quantity of a substance is the mole.

mole – this is such an amount of a substance that contains as many structural units of a given substance (molecules, atoms, ions, etc.) as there are carbon atoms in 0.012 kg (12 g) of a carbon isotope 12 FROM .

Given that the value of the absolute atomic mass for carbon is m(C) \u003d 1.99 10  26 kg, you can calculate the number of carbon atoms N BUT contained in 0.012 kg of carbon.

A mole of any substance contains the same number of particles of this substance (structural units). The number of structural units contained in a substance with an amount of one mole is 6.02 10 23 and called Avogadro's number (N BUT ).

For example, one mole of copper contains 6.02 10 23 copper atoms (Cu), and one mole of hydrogen (H 2) contains 6.02 10 23 hydrogen molecules.

molar mass(M) is the mass of a substance taken in an amount of 1 mol.

The molar mass is denoted by the letter M and has the unit [g/mol]. In physics, the dimension [kg/kmol] is used.

In the general case, the numerical value of the molar mass of a substance numerically coincides with the value of its relative molecular (relative atomic) mass.

For example, the relative molecular weight of water is:

Mr (H 2 O) \u003d 2Ar (H) + Ar (O) \u003d 2 ∙ 1 + 16 \u003d 18 a.m.u.

The molar mass of water has the same value, but is expressed in g/mol:

M (H 2 O) = 18 g/mol.

Thus, a mole of water containing 6.02 10 23 water molecules (respectively 2 6.02 10 23 hydrogen atoms and 6.02 10 23 oxygen atoms) has a mass of 18 grams. 1 mole of water contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms.

1.3.4. The relationship between the mass of a substance and its quantity

Knowing the mass of a substance and its chemical formula, and hence the value of its molar mass, one can determine the amount of a substance and, conversely, knowing the amount of a substance, one can determine its mass. For such calculations, you should use the formulas:

where ν is the amount of substance, [mol]; m is the mass of the substance, [g] or [kg]; M is the molar mass of the substance, [g/mol] or [kg/kmol].

For example, to find the mass of sodium sulfate (Na 2 SO 4) in the amount of 5 mol, we find:

1) the value of the relative molecular weight of Na 2 SO 4, which is the sum of the rounded values of the relative atomic masses:

Mr (Na 2 SO 4) \u003d 2Ar (Na) + Ar (S) + 4Ar (O) \u003d 142,

2) the value of the molar mass of the substance numerically equal to it:

M (Na 2 SO 4) = 142 g/mol,

3) and, finally, a mass of 5 mol of sodium sulfate:

m = ν M = 5 mol 142 g/mol = 710 g

Answer: 710.

1.3.5. The relationship between the volume of a substance and its quantity

Under normal conditions (n.o.), i.e. at pressure R , equal to 101325 Pa (760 mm Hg), and temperature T, equal to 273.15 K (0 С), one mole of various gases and vapors occupies the same volume, equal to 22.4 l.

The volume occupied by 1 mole of gas or vapor at n.o. is called molar volumegas and has the dimension of a liter per mole.

V mol \u003d 22.4 l / mol.

Knowing the amount of gaseous substance (ν ) and molar volume value (V mol) you can calculate its volume (V) under normal conditions:

V = ν V mol,

where ν is the amount of substance [mol]; V is the volume of the gaseous substance [l]; V mol \u003d 22.4 l / mol.

Conversely, knowing the volume ( V) of a gaseous substance under normal conditions, you can calculate its amount (ν) :

Molar and molal concentrations, despite similar names, are different values. Their main difference is that when determining the molal concentration, the calculation is made not on the volume of the solution, as in the detection of molarity, but on the mass of the solvent.

General information about solutions and solubility

A homogeneous system is called, which includes a number of components that are independent of each other. One of them is considered a solvent, and the rest are substances dissolved in it. A solvent is the substance that has the most in solution.

Solubility - the ability of a substance to form homogeneous systems with other substances - solutions in which it is in the form of individual atoms, ions, molecules or particles. Concentration is a measure of solubility.

Therefore, solubility is the ability of substances to be distributed evenly in the form of elementary particles throughout the volume of the solvent.

True solutions are classified as follows:

by type of solvent - non-aqueous and aqueous;
by type of solute - solutions of gases, acids, alkalis, salts, etc.;
on interaction with electric current - electrolytes (substances that have electrical conductivity) and non-electrolytes (substances that are not capable of electrical conductivity);
by concentration - diluted and concentrated.

Concentration and ways of expressing it

Concentration is the content (weight) of a substance dissolved in a certain amount (weight or volume) of the solvent or in a certain volume of the entire solution. It is of the following types:

1. Percentage concentration (expressed in%) - it tells how many grams of a solute are contained in 100 grams of a solution.

2. Molar concentration is the number of gram moles per 1 liter of solution. Shows how many gram molecules are contained in 1 liter of a substance solution.

3. Normal concentration is the number of gram equivalents per 1 liter of solution. Shows how many gram equivalents of a solute are contained in 1 liter of solution.

4. The molar concentration shows how much of the solute in moles falls on 1 kilogram of the solvent.

5. Titer determines the content (in grams) of a substance that is dissolved in 1 milliliter of solution.

Molar and molal concentrations are different from each other. Consider their individual characteristics.

Molar concentration

Formula to determine it:

Cv=(v/V), where

V is the total volume of the solution, liter or m 3.

For example, the entry "0.1 M solution of H 2 SO 4" indicates that 0.1 mol (9.8 grams) of sulfuric acid is present in 1 liter of such a solution.

Molar concentration

It should always be taken into account that molar and molar concentrations have completely different meanings.

What is the molal Formula for its definition is as follows:

Cm=(v/m), where

v is the amount of dissolved substance, mol;

m is the mass of the solvent, kg.

For example, writing a 0.2 M NaOH solution means that 0.2 mol of NaOH is dissolved in 1 kilogram of water (in this case, it is a solvent).

Additional formulas required for calculations

A lot of supporting information may be required in order for the molal concentration to be calculated. Formulas that can be useful for solving basic problems are presented below.

Under the amount of matter ν understand a certain number of atoms, electrons, molecules, ions or other particles.

v=m/M=N/N A =V/V m , where:

m is the mass of the compound, g or kg;
M - molar mass, g (or kg) / mol;
N is the number of structural units;
N A is the number of structural units in 1 mole of a substance, Avogadro's constant: 6.02. 10 23 mol - 1;
V is the total volume, l or m 3 ;
V m - molar volume, l / mol or m 3 / mol.

The latter is calculated by the formula:

V m =RT/P, where

R - constant, 8.314 J / (mol. K);
T - gas temperature, K;
P - gas pressure, Pa.

Examples of tasks for molarity and molality. Task #1

Determine the molar concentration of potassium hydroxide in a 500 ml solution. The mass of KOH in solution is 20 grams.

Definition

The molar mass of potassium hydroxide is:

M KOH \u003d 39 + 16 + 1 \u003d 56 g / mol.

We calculate how much is contained in the solution:

ν(KOH) \u003d m / M \u003d 20/56 \u003d 0.36 mol.

We take into account that the volume of the solution should be expressed in liters:

500 ml = 500/1000 = 0.5 liters.

Determine the molar concentration of potassium hydroxide:

Cv (KOH) \u003d v (KOH) / V (KOH) \u003d 0.36 / 0.5 \u003d 0.72 mol / liter.

Task #2

How much sulfur oxide (IV) under normal conditions (i.e. when P = 101325 Pa, and T = 273 K) should be taken in order to prepare a solution of sulfurous acid with a concentration of 2.5 mol / liter with a volume of 5 liters?

Definition

Determine how much is contained in the solution:

ν (H 2 SO 3) \u003d Cv (H 2 SO 3) ∙ V (solution) \u003d 2.5 ∙ 5 \u003d 12.5 mol.

The sulfuric acid production equation is as follows:

SO 2 + H 2 O \u003d H 2 SO 3

According to this:

ν(SO 2) \u003d ν(H 2 SO 3);

ν(SO 2) \u003d 12.5 mol.

Keeping in mind that under normal conditions, 1 mole of gas has a volume of 22.4 liters, we calculate the volume of sulfur oxide:

V (SO 2) \u003d ν (SO 2) ∙ 22.4 \u003d 12.5 ∙ 22.4 \u003d 280 liters.

Task #3

Determine the molar concentration of NaOH in the solution when it is equal to 25.5%, and the density is 1.25 g/ml.

Definition

We take as a sample a solution with a volume of 1 liter and determine its mass:

m (solution) = V (solution) ∙ p (solution) = 1000 ∙ 1.25 = 1250 grams.

We calculate how much alkali is in the sample by mass:

m (NaOH) \u003d (w ∙ m (solution)) / 100% \u003d (25.5 ∙ 1250) / 100 \u003d 319 grams.

Sodium hydroxide is equal to:

We calculate how much is contained in the sample:

v(NaOH) \u003d m / M \u003d 319/40 \u003d 8 mol.

Determine the molar concentration of alkali:

Cv (NaOH) \u003d v / V \u003d 8/1 \u003d 8 mol / liter.

Task #4

10 grams of NaCl salt were dissolved in water (100 grams). Set the concentration of the solution (molal).

Definition

The molar mass of NaCl is:

M NaCl \u003d 23 + 35 \u003d 58 g / mol.

The amount of NaCl contained in the solution:

ν(NaCl) \u003d m / M \u003d 10/58 \u003d 0.17 mol.

In this case, the solvent is water:

100 grams of water \u003d 100/1000 \u003d 0.1 kg of H 2 O in this solution.

The molar concentration of the solution will be equal to:

Cm (NaCl) \u003d v (NaCl) / m (water) \u003d 0.17 / 0.1 \u003d 1.7 mol / kg.

Task #5

Determine the molar concentration of a 15% NaOH alkali solution.

Definition

A 15% alkali solution means that every 100 grams of solution contains 15 grams of NaOH and 85 grams of water. Or that in every 100 kilograms of solution there are 15 kilograms of NaOH and 85 kilograms of water. In order to prepare it, it is necessary to dissolve 15 grams (kilogram) of alkali in 85 grams (kilograms) of H 2 O.

The molar mass of sodium hydroxide is:

M NaOH = 23 + 16 + 1 = 40 g/mol.

Now we find the amount of sodium hydroxide in the solution:

ν=m/M=15/40=0.375 mol.

Mass of solvent (water) in kilograms:

85 grams of H 2 O \u003d 85/1000 \u003d 0.085 kg of H 2 O in this solution.

After that, the molar concentration is determined:

Cm=(ν/m)=0.375/0.085=4.41 mol/kg.

In accordance with these typical tasks, it is possible to solve most of the others to determine molality and molarity.

1 mol per liter [mol/l] = 1000 mol per meter³ [mol/m³]

Initial value

Converted value

moles per meter³ moles per liter moles per centimeter³ moles per millimeters decimeter molar millimolar micromolar nanomolar picomolar femtomolar attomolar zeptomolar yoctomolar

More about molar concentration

General information

Factors affecting molar concentration

Units

How to find molar concentration

Examples

Na: 22.989769
H: 1.00794
C: 12.0107
O:15.9994

Application

First, we solve the equation for the chemical reaction between soda and acetic acid:

NaHCO₃ + CH₃COOH → NaC₂H₃O₂ + H₂CO₃

The reaction product is H₂CO₃, a substance that, due to low stability, enters into a chemical reaction again.

H₂CO₃ → H₂O + CO₂