The molar volume of a gas is a unit of measurement. Molar volume of gas

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 formed from Latin name 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 highest degree oxidation 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 covalentsconnectionher- 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. In the method of valence bonds, it is assumed 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 N 2 O 4 type dimers at the expense of 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.

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 With .

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) :

The volume of 1 mol of a substance is called the Molar volume. The molar mass of 1 mol of water = 18 g/mol 18 g of water occupies a volume of 18 ml. So the molar volume of water is 18 ml. 18 g of water occupy a volume equal to 18 ml, because. the density of water is 1 g/ml CONCLUSION: The molar volume depends on the density of the substance (for liquids and solids).

1 mole of any gas under normal conditions occupies the same volume equal to 22.4 liters. Normal conditions and their designations n.o.s. (0 0 С and 760 mm Hg; 1 atm.; 101.3 kPa). The volume of gas by the amount of substance 1 mol is called the molar volume and denoted - V m

Problem solving Problem 1 Given: V(NH 3) n.o.s. \u003d 33.6 m 3 Find: m -? Solution: 1. Calculate the molar mass of ammonia: M (NH 3) \u003d \u003d 17 kg / kmol

CONCLUSIONS 1. The volume of 1 mol of a substance is called the molar volume V m 2. For liquid and solid substances, the molar volume depends on their density 3. V m = 22.4 l / mol 4. Normal conditions (n.o.): and pressure 760 mm Hg, or 101.3 k Pa 5. The molar volume of gaseous substances is expressed in l / mol, ml / mmol,

In order to find out the composition of any gaseous substances, it is necessary to be able to operate with such concepts as molar volume, molar mass and the density of matter. In this article, we will consider what is molar volume, and how to calculate it?

Amount of substance

Quantitative calculations are carried out in order to actually carry out a particular process or find out the composition and structure of a certain substance. These calculations are inconvenient to make with the absolute values ​​of the masses of atoms or molecules due to the fact that they are very small. Relative atomic masses are also in most cases impossible to use, since they are not related to generally accepted measures of the mass or volume of a substance. Therefore, the concept of the amount of substance was introduced, which is denoted by the Greek letter v (nu) or n. The amount of a substance is proportional to the number of structural units (molecules, atomic particles) contained in the substance.

The unit of quantity of a substance is the mole.

A mole is the amount of a substance that contains as many structural units as there are atoms in 12 g of a carbon isotope.

The mass of 1 atom is 12 a. e. m., so the number of atoms in 12 g of the carbon isotope is:

Na \u003d 12g / 12 * 1.66057 * 10 to the power of -24g \u003d 6.0221 * 10 to the power of 23

The physical quantity Na is called the Avogadro constant. One mole of any substance contains 6.02 * 10 to the power of 23 particles.

Rice. 1. Avogadro's law.

Molar volume of gas

The molar volume of a gas is the ratio of the volume of a substance to the amount of that substance. This value is calculated by dividing the molar mass of a substance by its density according to the following formula:

where Vm is the molar volume, M is the molar mass, and p is the density of the substance.

Rice. 2. Molar volume formula.

AT international system Si measurement of the molar volume of gaseous substances is carried out in cubic meters per mol (m 3 / mol)

The molar volume of gaseous substances differs from substances in liquid and solid states in that a gaseous element with an amount of 1 mol always occupies the same volume (if the same parameters are observed).

The volume of gas depends on temperature and pressure, so the calculation should take the volume of gas under normal conditions. Normal conditions are considered to be a temperature of 0 degrees and a pressure of 101.325 kPa. The molar volume of 1 mol of gas under normal conditions is always the same and equal to 22.41 dm 3 /mol. This volume is called the molar volume of an ideal gas. That is, in 1 mole of any gas (oxygen, hydrogen, air), the volume is 22.41 dm 3 / m.

Rice. 3. Molar volume of gas under normal conditions.

Table "molar volume of gases"

The following table shows the volume of some gases:

What have we learned?

The molar volume of a gas studied in chemistry (grade 8), along with the molar mass and density, are the necessary quantities to determine the composition of one or another chemical. A feature of a molar gas is that one mole of gas always contains the same volume. This volume is called the molar volume of the gas.

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The volume of a gram-molecule of a gas, as well as the mass of a gram-molecule, is a derived unit of measurement and is expressed as the ratio of units of volume - liters or milliliters to a mole. Therefore, the dimension of the gram-molecular volume is l / mol or ml / mol. Since the volume of a gas depends on temperature and pressure, the gram-molecular volume of a gas varies depending on the conditions, but since the gram-molecules of all substances contain the same number of molecules, the gram-molecules of all substances under the same conditions occupy the same volume. under normal conditions. = 22.4 l/mol, or 22400 ml/mol. Recalculation of the gram-molecular volume of gas under normal conditions per volume under given conditions of production. is calculated according to the equation: J-t-tr from which it follows that where Vo is the gram-molecular volume of gas under normal conditions, Umol is the desired gram-molecular volume of gas. Example. Calculate the gram-molecular volume of the gas at 720 mm Hg. Art. and 87°C. Decision. The most important calculations related to the gram-molecular volume of a gas a) Converting the volume of gas to the number of moles and the number of moles per volume of gas. Example 1. Calculate how many moles are contained in 500 liters of gas under normal conditions. Decision. Example 2. Calculate the volume of 3 mol of gas at 27 * C 780 mm Hg. Art. Decision. We calculate the gram-molecular volume of gas under the specified conditions: V - ™ ** RP st. - 22.A l / mol. 300 deg \u003d 94 p. -273 vrad 780 mm Hg "ap.--24" ° Calculate the volume of 3 mol GRAM MOLECULAR VOLUME OF GAS V \u003d 24.0 l / mol 3 mol \u003d 72 l b) Conversion of the mass of gas to its volume and volume of a gas per its mass. In the first case, the number of moles of gas is first calculated from its mass, and then the volume of gas is calculated from the found number of moles. In the second case, the number of moles of gas is first calculated from its volume, and then, from the found number of moles, the mass of the gas. Example 1, Calculate the volume (at N.C.) of 5.5 g of carbon dioxide CO * Solution. |icoe ■= 44 g/mol V = 22.4 l/mol 0.125 mol 2.80 l Example 2. Calculate the mass of 800 ml (at n.a.) carbon monoxide CO. Decision. | * w => 28 g / mol m " 28 g / lnm 0.036 did * \u003d" 1.000 g If the mass of the gas is expressed not in grams, but in kilograms or tons, and its volume is expressed not in liters or milliliters, but in cubic meters , then a twofold approach to these calculations is possible: either split higher measures into lower ones, or the calculation of ae with moles is known, and with kilogram-molecules or ton-molecules, using the following ratios: under normal conditions, 1 kilogram-molecule-22,400 l / kmol , 1 ton-molecule - 22,400 m*/tmol. Units: kilogram-molecule - kg/kmol, ton-molecule - t/tmol. Example 1. Calculate the volume of 8.2 tons of oxygen. Decision. 1 ton-molecule Oa » 32 t/tmol. We find the number of ton-molecules of oxygen contained in 8.2 tons of oxygen: 32 t/tmol ** 0.1 Calculate the mass of 1000 -k * ammonia (at n.a.). Decision. We calculate the number of ton-molecules in the specified amount of ammonia: "-stay5JT-0.045 t/mol Calculate the mass of ammonia: 1 ton-molecule NH, 17 t/mol tyv, \u003d 17 t/mol 0.045 t/mol * 0.765 t General principle of calculation, related to gas mixtures, is that the calculations related to the individual components are performed separately, and then the results are summed.Example 1. Calculate what volume a gas mixture consisting of 140 g of nitrogen and 30 e of hydrogen will occupy under normal conditions. Solution Calculate the number of moles of nitrogen and hydrogen contained in the mixture (No. "= 28 u/mol; cn, = 2 g/mol): 140 £ 30 in 28 g/mol W Total 20 mol. GRAM MOLECULAR VOLUME OF GAS Calculate the volume of the mixture : Ueden in 22 "4 AlnoAb 20 mol " 448 l Example 2. Calculate the mass of 114 mixture (at n.a.) of carbon monoxide and carbon dioxide, the volume composition of which is expressed by the ratio: /lso: /iso, = 8:3. Decision. According to the indicated composition, we find the volumes of each gas by the method of proportional division, after which we calculate the corresponding number of moles: t / II l "8 Q" "11 J 8 Q Ksoe 8 + 3 8 * Va> "a & + & * VCQM grfc - 0 "36 ^-grfc "" 0.134 jas * Calculating! the mass of each of the gases from the found number of moles of each of them. 1 "co 28 g / mol; jico. \u003d 44 g / mol moo "28 e! mol 0.36 mol "South tco. \u003d 44 e / zham" - 0.134 "au> - 5.9 g By adding the found masses of each of the components, we find the mass of the mixture: gas by gram-molecular volume Above was considered the method of calculating the molecular weight of a gas by relative density.Now we will consider the method of calculating the molecular weight of a gas by gram-molecular volume.In the calculation, it is assumed that the mass and volume of the gas are directly proportional to each other.It follows "that the volume of a gas and its mass are related to each other as the gram-molecular volume of a gas is to the gram- molecular weight it, which is expressed in mathematical form as follows: V_ Ushts / i (x where Yn * "- gram-molecular volume, p - gram-molecular weight. From here _ Uiol t p? Consider the calculation method using a specific example. "Example. Mass 34 $ ju gas at 740 mmHg, spi and 21°C is equal to 0.604 g Calculate the molecular weight of the gas Solution To solve, you need to know the gram-molecular volume of the gas. then a certain gram-molecular volume of gas.You can use the standard gram-molecular volume of gas, which is equal to 22.4 l / mol. Then the volume of gas indicated in the condition of the problem should be reduced to normal conditions. But it is possible, on the contrary, to calculate the gram-molecular volume of a gas under the conditions specified in the problem. With the first method of calculation, the following design is obtained: at 740 * mrt.st .. 340 ml - 273 deg ^ Q ^ 0 760 mm Hg. Art. 294 deg ™ 1 l.1 - 22.4 l / mol 0.604 in _ s, ypya. -m-8 \u003d 44 g, M0Ab In the second method, we find: V - 22»4 A! mol No. mm Hg. st.-29A deg 0A77 l1ylv. Uiol 273 vrad 740 mmHg Art. ~ R * 0 ** In both cases, we calculate the mass of the gram molecule, but since the gram molecule is numerically equal to the molecular weight, we thereby find the molecular weight.