Where m is mass, M is molar mass, V is volume.

4. Law of Avogadro. Established by the Italian physicist Avogadro in 1811. The same volumes of any gases, taken at the same temperature and the same pressure, contain the same number of molecules.

Thus, the concept of the amount of a substance can be formulated: 1 mole of a substance contains a number of particles equal to 6.02 * 10 23 (called the Avogadro constant)

The consequence of this law is that 1 mole of any gas occupies under normal conditions (P 0 \u003d 101.3 kPa and T 0 \u003d 298 K) a volume equal to 22.4 liters.

5. Boyle-Mariotte Law

At constant temperature, the volume of a given amount of gas is inversely proportional to the pressure under which it is:

6. Gay-Lussac's law

At constant pressure, the change in the volume of a gas is directly proportional to the temperature:

V/T = const.

7. The relationship between gas volume, pressure and temperature can be expressed the combined law of Boyle-Mariotte and Gay-Lussac, which is used to bring gas volumes from one condition to another:

P 0 , V 0 ,T 0 - volume pressure and temperature under normal conditions: P 0 =760 mm Hg. Art. or 101.3 kPa; T 0 \u003d 273 K (0 0 C)

8. Independent assessment of the value of molecular masses M can be done using the so-called equations of state for an ideal gas or the Clapeyron-Mendeleev equations :

pV=(m/M)*RT=vRT.(1.1)

Where R - gas pressure in a closed system, V- volume of the system, T - mass of gas T - absolute temperature, R- universal gas constant.

Note that the value of the constant R can be obtained by substituting the values ​​characterizing one mole of gas at N.C. into equation (1.1):

r = (p V) / (T) \u003d (101.325kPa 22.4 l) / (1 mol 273K) \u003d 8.31J / mol.K)

Examples of problem solving

Example 1 Bringing the volume of gas to normal conditions.

What volume (n.o.) will occupy 0.4×10 -3 m 3 of gas at 50 0 C and a pressure of 0.954×10 5 Pa?

Solution. To bring the volume of gas to normal conditions, use the general formula that combines the laws of Boyle-Mariotte and Gay-Lussac:

pV/T = p 0 V 0 /T 0 .

The volume of gas (n.o.) is, where T 0 \u003d 273 K; p 0 \u003d 1.013 × 10 5 Pa; T = 273 + 50 = 323 K;

M 3 \u003d 0.32 × 10 -3 m 3.

When (n.o.) gas occupies a volume equal to 0.32×10 -3 m 3 .

Example 2 Calculation of the relative density of a gas from its molecular weight.

Calculate the density of ethane C 2 H 6 from hydrogen and air.

Solution. It follows from Avogadro's law that the relative density of one gas over another is equal to the ratio of molecular masses ( M h) of these gases, i.e. D=M 1 /M 2. If M 1С2Н6 = 30, M 2 H2 = 2, the average molecular weight of air is 29, then the relative density of ethane with respect to hydrogen is D H2 = 30/2 =15.

Relative density of ethane in air: D air= 30/29 = 1.03, i.e. ethane is 15 times heavier than hydrogen and 1.03 times heavier than air.

Example 3 Determination of the average molecular weight of a mixture of gases by relative density.

Calculate the average molecular weight of a mixture of gases consisting of 80% methane and 20% oxygen (by volume) using the values ​​of the relative density of these gases with respect to hydrogen.

Solution. Often calculations are made according to the mixing rule, which is that the ratio of the volumes of gases in a two-component gas mixture is inversely proportional to the differences between the density of the mixture and the densities of the gases that make up this mixture. Let us denote the relative density of the gas mixture with respect to hydrogen through D H2. it will be greater than the density of methane, but less than the density of oxygen:

80D H2 - 640 = 320 - 20 D H2; D H2 = 9.6.

The hydrogen density of this mixture of gases is 9.6. average molecular weight of the gas mixture M H2 = 2 D H2 = 9.6×2 = 19.2.

Example 4 Calculation of the molar mass of a gas.

The mass of 0.327 × 10 -3 m 3 of gas at 13 0 C and a pressure of 1.040 × 10 5 Pa is 0.828 × 10 -3 kg. Calculate the molar mass of the gas.

Solution. You can calculate the molar mass of a gas using the Mendeleev-Clapeyron equation:

Where m is the mass of gas; M is the molar mass of the gas; R- molar (universal) gas constant, the value of which is determined by the accepted units of measurement.

If the pressure is measured in Pa, and the volume in m 3, then R\u003d 8.3144 × 10 3 J / (kmol × K).

3.1. When performing measurements of atmospheric air, air of the working area, as well as industrial emissions and hydrocarbons in gas pipelines, there is a problem of bringing the volumes of measured air to normal (standard) conditions. Often in practice, when conducting air quality measurements, the conversion of measured concentrations to normal conditions is not used, resulting in unreliable results.

Here is an excerpt from the Standard:

“Measurements are brought to standard conditions using the following formula:

C 0 \u003d C 1 * P 0 T 1 / R 1 T 0

where: C 0 - the result, expressed in units of mass per unit volume of air, kg / cu. m, or the amount of substance per unit volume of air, mol / cu. m, at standard temperature and pressure;

C 1 - the result, expressed in units of mass per unit volume of air, kg / cu. m, or the amount of substance per unit volume

air, mol/cu. m, at temperature T 1, K, and pressure P 1, kPa.

The formula for bringing to normal conditions in a simplified form has the form (2)

C 1 \u003d C 0 * f, where f \u003d P 1 T 0 / P 0 T 1

standard conversion factor for normalization. The parameters of air and impurities are measured at different temperatures, pressures and humidity. The results lead to standard conditions for comparing measured air quality parameters in different locations and different climates.

3.2 Industry normal conditions

Normal conditions are the standard physical conditions with which the properties of substances are usually correlated (Standard temperature and pressure, STP). Normal conditions are defined by IUPAC (International Union of Practical and Applied Chemistry) as follows: Atmospheric pressure 101325 Pa = 760 mm Hg. Air temperature 273.15 K = 0° C.

Standard conditions (Standard Ambient Temperature and Pressure, SATP) are normal ambient temperature and pressure: pressure 1 Bar = 10 5 Pa = 750.06 mm T. St.; temperature 298.15 K = 25 °C.

Other areas.

Air quality measurements.

The results of measurements of concentrations of harmful substances in the air of the working area lead to the following conditions: a temperature of 293 K (20°C) and a pressure of 101.3 kPa (760 mm Hg).

Aerodynamic parameters of pollutant emissions must be measured in accordance with current state standards. The volumes of exhaust gases obtained from the results of instrumental measurements must be brought to normal conditions (n.s.): 0 ° C, 101.3 kPa ..

Aviation.

The International Civil Aviation Organization (ICAO) defines the International Standard Atmosphere (ISA) at sea level with a temperature of 15°C, an atmospheric pressure of 101325 Pa, and a relative humidity of 0%. These parameters are used when calculating the movement of aircraft.

Gas economy.

The gas industry of the Russian Federation uses atmospheric conditions in accordance with GOST 2939-63 for settlements with consumers: temperature 20°C (293.15K); pressure 760 mm Hg. Art. (101325 N/m²); humidity is 0. Thus, the mass of a cubic meter of gas according to GOST 2939-63 is somewhat less than under “chemical” normal conditions.

Tests

For testing machines, instruments and other technical products, the following are taken as normal values ​​of climatic factors when testing products (normal climatic test conditions):

Temperature - plus 25°±10°С; Relative humidity - 45-80%

Atmospheric pressure 84-106 kPa (630-800 mmHg)

Verification of measuring instruments

The nominal values ​​of the most common normal influencing quantities are selected as follows: Temperature - 293 K (20°C), atmospheric pressure - 101.3 kPa (760 mmHg).

Rationing

The guidelines for setting air quality standards indicate that MPCs in ambient air are set under normal indoor conditions, i.e. 20 C and 760 mm. rt. Art.

The molar volume of a gas is equal to the ratio of the volume of gas to the amount of substance of this gas, i.e.

V m = V(X) / n(X),

where V m - molar volume of gas - a constant value for any gas under given conditions;

V(X) is the volume of gas X;

n(X) is the amount of gas substance X.

The molar volume of gases under normal conditions (normal pressure p n \u003d 101 325 Pa ≈ 101.3 kPa and temperature T n \u003d 273.15 K ≈ 273 K) is V m \u003d 22.4 l / mol.

Laws of ideal gases

In calculations involving gases, it is often necessary to switch from these conditions to normal conditions or vice versa. In this case, it is convenient to use the formula following from the combined gas law of Boyle-Mariotte and Gay-Lussac:

pV / T = p n V n / T n

Where p is the pressure; V - volume; T is the temperature on the Kelvin scale; the index "n" indicates normal conditions.

Volume fraction

The composition of gas mixtures is often expressed using a volume fraction - the ratio of the volume of a given component to the total volume of the system, i.e.

φ(X) = V(X) / V

where φ(X) - volume fraction of component X;

V(X) - volume of component X;

V is the volume of the system.

The volume fraction is a dimensionless quantity, it is expressed in fractions of a unit or as a percentage.

Example 1. What volume will take at a temperature of 20 ° C and a pressure of 250 kPa ammonia weighing 51 g?

1. Determine the amount of ammonia substance: n (NH 3) \u003d m (NH 3) / M (NH 3) \u003d 51/17 \u003d 3 mol. 2. The volume of ammonia under normal conditions is: V (NH 3) \u003d V m n (NH 3) \u003d 22.4 3 \u003d 67.2 l. 3. Using formula (3), we bring the volume of ammonia to these conditions (temperature T = (273 + 20) K = 293 K): V (NH 3) \u003d p n V n (NH 3) / pT n \u003d 101.3 293 67.2 / 250 273 \u003d 29.2 l. Answer: V (NH 3) \u003d 29.2 liters.

Example 2. Determine the volume that a gas mixture containing hydrogen, weighing 1.4 g and nitrogen, weighing 5.6 g, will take under normal conditions.

1. Find the amount of hydrogen and nitrogen matter: n (N 2) \u003d m (N 2) / M (N 2) \u003d 5.6 / 28 \u003d 0.2 mol n (H 2) \u003d m (H 2) / M (H 2) \u003d 1.4 / 2 \u003d 0.7 mol 2. Since under normal conditions these gases do not interact with each other, the volume of the gas mixture will be equal to the sum of the volumes of gases, i.e. V (mixtures) \u003d V (N 2) + V (H 2) \u003d V m n (N 2) + V m n (H2) \u003d 22.4 0.2 + 22.4 0.7 \u003d 20.16 l. Answer: V (mixture) \u003d 20.16 liters.

Law of Volumetric Relations

How to solve the problem using the "Law of Volumetric Relations"?

Law of volumetric ratios: The volumes of gases involved in a reaction are related to each other as small integers equal to the coefficients in the reaction equation.

The coefficients in the reaction equations show the number of volumes of reacting and formed gaseous substances.

Example. Calculate the volume of air required to burn 112 liters of acetylene.

1. We compose the reaction equation:

2. Based on the law of volumetric ratios, we calculate the volume of oxygen:

112/2 \u003d X / 5, whence X \u003d 112 5 / 2 \u003d 280l

3. Determine the volume of air:

V (air) \u003d V (O 2) / φ (O 2)

V (air) \u003d 280 / 0.2 \u003d 1400 l.

: V \u003d n * Vm, where V is the volume of gas (l), n is the amount of substance (mol), Vm is the molar volume of gas (l / mol), at normal (n.o.) is a standard value and is equal to 22, 4 l/mol. It happens that in the condition there is no amount of a substance, but there is a mass of a certain substance, then we do this: n = m / M, where m is the mass of the substance (g), M is the molar mass of the substance (g / mol). We find the molar mass according to the table D.I. Mendeleev: under each element is its atomic mass, add up all the masses and get the one we need. But such tasks are quite rare, usually there is a . The solution to such problems is slightly different. Let's look at an example.

What volume of hydrogen will be released under normal conditions if aluminum weighing 10.8 g is dissolved in an excess of hydrochloric acid.

If we are dealing with a gas system, then the following formula takes place: q(x) = V(x)/V, where q(x)(phi) is the fraction of the component, V(x) is the volume of the component (l), V is the volume of the system (l). To find the volume of the component, we obtain the formula: V(x) = q(x)*V. And if you need to find the volume of the system, then: V = V(x)/q(x).

note

There are other formulas for finding the volume, but if you need to find the volume of a gas, only the formulas given in this article will do.

Sources:

• "Manual in Chemistry", G.P. Khomchenko, 2005.
• how to find scope of work
• Find the volume of hydrogen in the electrolysis of a solution of ZnSO4

An ideal gas is one in which the interaction between molecules is negligible. In addition to pressure, the state of a gas is characterized by temperature and volume. The relationships between these parameters are displayed in the gas laws.

Instruction

The pressure of a gas is directly proportional to its temperature, the amount of substance, and inversely proportional to the volume of the vessel occupied by the gas. The coefficient of proportionality is the universal gas constant R, approximately equal to 8.314. It is measured in joules divided by moles and by.

This provision forms the mathematical dependence P=νRT/V, where ν is the amount of substance (mol), R=8.314 is the universal gas constant (J/mol K), T is the gas temperature, V is the volume. The pressure is expressed in . It can be expressed and, while 1 atm \u003d 101.325 kPa.

The considered dependence is a consequence of the Mendeleev-Clapeyron equation PV=(m/M) RT. Here m is the mass of the gas (g), M is its molar mass (g / mol), and the fraction m / M gives as a result the amount of substance ν, or the number of moles. The Mendeleev-Clapeyron equation is valid for all gases that can be considered. This is a physical gas law.

Names of acids are formed from the Russian name of the central acid atom with the addition of suffixes and endings. If the oxidation state of the central atom of the acid corresponds to the group number of the Periodic system, then the name is formed using the simplest adjective from the name of the element: H 2 SO 4 - sulfuric acid, HMnO 4 - manganese acid. If acid-forming elements have two oxidation states, then the intermediate oxidation state is indicated by the suffix -ist-: H 2 SO 3 - sulfurous acid, HNO 2 - nitrous acid. For the names of halogen acids with many oxidation states, various suffixes are used: typical examples - HClO 4 - chlorine n th acid, HClO 3 - chlorine novat th acid, HClO 2 - chlorine ist acid, HClO - chlorine novatist acid (the anoxic acid HCl is called hydrochloric acid—usually hydrochloric acid). Acids can differ in the number of water molecules that hydrate the oxide. Acids containing the largest number of hydrogen atoms are called ortho acids: H 4 SiO 4 - orthosilicic acid, H 3 PO 4 - phosphoric acid. Acids containing 1 or 2 hydrogen atoms are called metaacids: H 2 SiO 3 - metasilicic acid, HPO 3 - metaphosphoric acid. Acids containing two central atoms are called di acids: H 2 S 2 O 7 - disulfuric acid, H 4 P 2 O 7 - diphosphoric acid.

The names of complex compounds are formed in the same way as salt names, but the complex cation or anion is given a systematic name, that is, it is read from right to left: K 3 - potassium hexafluoroferrate (III), SO 4 - tetraammine copper (II) sulfate.

Names of oxides are formed using the word "oxide" and the genitive case of the Russian name of the central oxide atom, indicating, if necessary, the degree of oxidation of the element: Al 2 O 3 - aluminum oxide, Fe 2 O 3 - iron oxide (III).

Base names are formed using the word "hydroxide" and the genitive case of the Russian name of the central hydroxide atom, indicating, if necessary, the degree of oxidation of the element: Al (OH) 3 - aluminum hydroxide, Fe (OH) 3 - iron (III) hydroxide.

Names of compounds with hydrogen are formed depending on the acid-base properties of these compounds. For gaseous acid-forming compounds with hydrogen, the names are used: H 2 S - sulfane (hydrogen sulfide), H 2 Se - selane (hydrogen selenide), HI - hydrogen iodine; their solutions in water are called, respectively, hydrosulfide, hydroselenic and hydroiodic acids. For some compounds with hydrogen, special names are used: NH 3 - ammonia, N 2 H 4 - hydrazine, PH 3 - phosphine. Compounds with hydrogen having an oxidation state of –1 are called hydrides: NaH is sodium hydride, CaH 2 is calcium hydride.

Names of salts are formed from the Latin name of the central atom of the acid residue with the addition of prefixes and suffixes. The names of binary (two-element) salts are formed using the suffix - id: NaCl - sodium chloride, Na 2 S - sodium sulfide. If the central atom of an oxygen-containing acid residue has two positive oxidation states, then the highest oxidation state is indicated by the suffix - at: Na 2 SO 4 - sulf at sodium, KNO 3 - nitr at potassium, and the lowest oxidation state - the suffix - it: Na 2 SO 3 - sulf it sodium, KNO 2 - nitr it potassium. For the name of oxygen-containing salts of halogens, prefixes and suffixes are used: KClO 4 - lane chlorine at potassium, Mg (ClO 3) 2 - chlorine at magnesium, KClO 2 - chlorine it potassium, KClO - hypo chlorine it potassium.

Saturation covalentsconnectionto her- manifests itself in the fact that there are no unpaired electrons in the compounds of s- and p-elements, that is, all unpaired electrons of atoms form bonding electron pairs (exceptions are NO, NO 2, ClO 2 and ClO 3).

Lone electron pairs (LEPs) are electrons that occupy atomic orbitals in pairs. The presence of NEP determines the ability of anions or molecules to form donor-acceptor bonds as donors of electron pairs.

Unpaired electrons - electrons of an atom, contained one by one in the orbital. For s- and p-elements, the number of unpaired electrons determines how many bonding electron pairs a given atom can form with other atoms by the exchange mechanism. The valence bond method assumes that the number of unpaired electrons can be increased by unshared electron pairs if there are vacant orbitals within the valence electronic level. In most compounds of s- and p-elements, there are no unpaired electrons, since all unpaired electrons of atoms form bonds. However, molecules with unpaired electrons exist, for example, NO, NO 2 , they are highly reactive and tend to form dimers of the N 2 O 4 type due to unpaired electrons.

Normal concentration - is the number of moles equivalents in 1 liter of solution.

Normal conditions - temperature 273K (0 o C), pressure 101.3 kPa (1 atm).

Exchange and donor-acceptor mechanisms of chemical bond formation. The formation of covalent bonds between atoms can occur in two ways. If the formation of a bonding electron pair occurs due to the unpaired electrons of both bonded atoms, then this method of formation of a bonding electron pair is called the exchange mechanism - the atoms exchange electrons, moreover, the bonding electrons belong to both bonded atoms. If the bonding electron pair is formed due to the lone electron pair of one atom and the vacant orbital of another atom, then such formation of the bonding electron pair is a donor-acceptor mechanism (see Fig. valence bond method).

Reversible ionic reactions - these are reactions in which products are formed that are capable of forming starting substances (if we keep in mind the written equation, then about reversible reactions we can say that they can proceed in both directions with the formation of weak electrolytes or poorly soluble compounds). Reversible ionic reactions are often characterized by incomplete conversion; since during a reversible ionic reaction, molecules or ions are formed that cause a shift towards the initial reaction products, that is, they “slow down” the reaction, as it were. Reversible ionic reactions are described using the ⇄ sign, and irreversible reactions are described using the → sign. An example of a reversible ionic reaction is the reaction H 2 S + Fe 2+ ⇄ FeS + 2H +, and an example of an irreversible one is S 2- + Fe 2+ → FeS.

Oxidizers – substances in which, during redox reactions, the oxidation states of some elements decrease.

Redox duality - the ability of substances to act redox reactions as an oxidizing agent or reducing agent, depending on the partner (for example, H 2 O 2 , NaNO 2).

Redox reactions(OVR) - These are chemical reactions during which the oxidation states of the elements of the reactants change.

Redox potential - a value that characterizes the redox ability (strength) of both the oxidizing agent and the reducing agent, which make up the corresponding half-reaction. Thus, the redox potential of the Cl 2 /Cl - pair, equal to 1.36 V, characterizes molecular chlorine as an oxidizing agent and chloride ion as a reducing agent.

Oxides - compounds of elements with oxygen, in which oxygen has an oxidation state of -2.

Orientation interactions– intermolecular interactions of polar molecules.

Osmosis - the phenomenon of the transfer of solvent molecules on a semi-permeable (solvent-permeable only) membrane towards a lower solvent concentration.

Osmotic pressure - physicochemical property of solutions, due to the ability of membranes to pass only solvent molecules. The osmotic pressure from the side of the less concentrated solution equalizes the penetration rates of the solvent molecules on both sides of the membrane. The osmotic pressure of a solution is equal to the pressure of a gas in which the concentration of molecules is the same as the concentration of particles in the solution.

Foundations according to Arrhenius - substances that, in the process of electrolytic dissociation, split off hydroxide ions.

Foundations according to Bronsted - compounds (molecules or ions such as S 2-, HS -) that can attach hydrogen ions.

Foundations according to Lewis (Lewis bases) – compounds (molecules or ions) with unshared electron pairs capable of forming donor-acceptor bonds. The most common Lewis base are water molecules, which have strong donor properties.

^ Molar mass and molar volume of a substance. Molar mass is the mass of a mole of a substance. It is calculated through the mass and amount of the substance according to the formula:

Mv \u003d K · Mr (1)

Where: K - coefficient of proportionality, equal to 1g / mol.

Indeed, for the carbon isotope 12 6 С Ar = 12, and the molar mass of atoms (according to the definition of the concept of “mol”) is 12 g / mol. Consequently, the numerical values ​​of the two masses are the same, and hence K = 1. It follows that the molar mass of a substance, expressed in grams per mole, has the same numerical value as its relative molecular weight(atomic) weight. Thus, the molar mass of atomic hydrogen is 1.008 g/mol, molecular hydrogen is 2.016 g/mol, and molecular oxygen is 31.999 g/mol.

According to Avogadro's law, the same number of molecules of any gas occupies the same volume under the same conditions. On the other hand, 1 mole of any substance contains (by definition) the same number of particles. It follows that at a certain temperature and pressure, 1 mole of any substance in the gaseous state occupies the same volume.

The ratio of the volume occupied by a substance to its quantity is called the molar volume of the substance. Under normal conditions (101.325 kPa; 273 K), the molar volume of any gas is 22,4l/mol(more precisely, Vn = 22.4 l/mol). This statement is true for such a gas, when other types of interaction of its molecules with each other, except for their elastic collision, can be neglected. Such gases are called ideal. For non-ideal gases, called real gases, the molar volumes are different and somewhat different from the exact value. However, in most cases, the difference affects only the fourth and subsequent significant figures.

Measurements of gas volumes are usually carried out under conditions other than normal. To bring the volume of gas to normal conditions, you can use the equation that combines the gas laws of Boyle - Mariotte and Gay - Lussac:

pV / T = p 0 V 0 / T 0

Where: V is the volume of gas at pressure p and temperature T;

V 0 is the volume of gas at normal pressure p 0 (101.325 kPa) and temperature T 0 (273.15 K).

The molar masses of gases can also be calculated using the equation of state of an ideal gas - the Clapeyron-Mendeleev equation:

pV = m B RT / M B ,

Where: p – gas pressure, Pa;

V is its volume, m 3;

M B - mass of substance, g;

M B is its molar mass, g/mol;

T is the absolute temperature, K;

R is the universal gas constant, equal to 8.314 J / (mol K).

If the volume and pressure of the gas are expressed in other units, then the value of the gas constant in the Clapeyron-Mendeleev equation will take on a different value. It can be calculated by the formula following from the combined law of the gaseous state for a mole of a substance under normal conditions for one mole of gas:

R = (p 0 V 0 / T 0)

Example 1 Express in moles: a) 6.0210 21 CO 2 molecules; b) 1.2010 24 oxygen atoms; c) 2.0010 23 water molecules. What is the molar mass of these substances?

Solution. A mole is the amount of a substance that contains the number of particles of any particular kind, equal to the Avogadro constant. Hence, a) 6.0210 21 i.e. 0.01 mol; b) 1.2010 24 , i.e. 2 mol; c) 2.0010 23 , i.e. 1/3 mol. The mass of a mole of a substance is expressed in kg/mol or g/mol. The molar mass of a substance in grams is numerically equal to its relative molecular (atomic) mass, expressed in atomic mass units (a.m.u.)

Since the molecular weights of CO 2 and H 2 O and the atomic mass of oxygen, respectively, are 44; 18 and 16 amu, then their molar masses are: a) 44 g/mol; b) 18 g/mol; c) 16g/mol.

Example 2 Calculate the absolute mass of the sulfuric acid molecule in grams.

Solution. A mole of any substance (see example 1) contains the Avogadro constant N A of structural units (in our example, molecules). The molar mass of H 2 SO 4 is 98.0 g/mol. Therefore, the mass of one molecule is 98/(6.02 10 23) = 1.63 10 -22 g.

Molar volume- the volume of one mole of a substance, the value obtained by dividing the molar mass by the density. Characterizes the packing density of molecules.

Meaning N A = 6.022…×10 23 It is called the Avogadro number after the Italian chemist Amedeo Avogadro. This is a universal constant for the smallest particles of any substance.

It is this number of molecules that contains 1 mole of oxygen O 2, the same number of atoms in 1 mole of iron (Fe), molecules in 1 mole of water H 2 O, etc.

According to Avogadro's law, 1 mole of an ideal gas at normal conditions has the same volume Vm\u003d 22.413 996 (39) l. Under normal conditions, most gases are close to ideal, so all reference information on the molar volume of chemical elements refers to their condensed phases, unless otherwise stated.