Task 3.1.1 How many moles of iron (II) sulfide are in 8.8 g of FeS?

Decision: Determine the molar mass (M) of iron (II) sulfide.

M (FeS) \u003d 56 + 32 \u003d 88 g / mol

Let's calculate how many moles are contained in 8.8 g of FeS

Task 3.1.2 How many molecules are there in 54 g of water? What is the mass of one water molecule?

Decision: Determine the molar mass of water.

M (H 2 O) \u003d 18 g / mol

Therefore, 54 g of water contains 54/18 \u003d 3 mol of H2O. One mol of any substance contains 6.02  10 23 molecules. Then 3 moles (54g H 2 O) contain 6.02  10 23  3 \u003d 18.06  10 23 molecules.

Let's determine the mass of one water molecule.

Task 3.1.3 How many moles and molecules are there in 1 m 3 of any gas under normal conditions?

Decision: 1 mole of any gas under normal conditions takes a volume of 22.4 liters. Therefore, 1m 3 (\u003d 1000l) will contain 44.6 moles of gas:

1 mole of any gas contains 6.02  10 23 molecules. From this it follows that 1 m 3 of any gas under normal conditions contains

6.02  10 23  44.6 \u003d 2.68  10 25 molecules.

Task 3.1.4 Express in moles: a) 6.02  10 23 molecules C 2 H 2;

b) 1.80  10 24 nitrogen atoms; c) 3.01  10 23 NH 3 molecules. What is the molar mass of these substances?

Decision: A mole is the amount of a substance that contains the number of particles of any particular kind, equal to Avogadro's constant. Hence:

and)

b)

in)

The mole mass of a substance is expressed in g / mol ... The molar mass of a substance in grams is numerically equal to its relative molecular (atomic) mass, expressed in amu.

Therefore, the molar masses of these substances are equal:

a) M (C 2 H 2) \u003d 26 g / mol; b) M (N) \u003d 14 g / mol;

c) M (NH 3) \u003d 17 g / mol;

Task 3.1.5 Determine the molar mass of the gas if, under normal conditions, 0.824 g is occupied by a volume of 0.260 liters.

Decision: Under normal conditions, 1 mole of any gas occupies a volume of 22.4 liters. Having calculated the mass of 22.4 liters of this gas, we find out its molar mass.

0.824 g of gas takes up a volume of 0.260 l

X g of gas take up a volume of 22.4 liters

Consequently, the molar mass of the gas is 71 g / mol.

3.2 Equivalent. Equivalence factor. Molar mass equivalents.

Task 3.2.1 Calculate the equivalent, equivalence factor and molar mass of equivalents H 3 RO 4 in metabolic reactions resulting in the formation of acidic and normal salts.

Decision Let us write down the reaction equations for the interaction of phosphoric acid with alkali:

H 2 PO 4 + NaOH \u003d NaH 2 PO 4 + H 2 O; (1)

H 2 PO 4 H 2 PO 4 + 2NaOH \u003d Na 2 HPO 4 + 2H 2 O; (2)

H 2 PO 4 + 3NaOH \u003d Na 3 PO 4 + 3H 2 O; (3)

Because phosphoric acid is a tribasic acid, it forms two acidic salts: NaH 2 PO 4 - sodium dihydrogen phosphate and Na 2 HPO 4 - sodium hydrogen phosphate and one average salt Na 3 PO 4 - sodium phosphate.

In reaction (1), phosphoric acid exchanges one hydrogen atom for a metal, i.e. behaves like a monobasic acid, therefore

f e (H 3 PO 4) in reaction (1) is equal to 1;

E (H 3 PO 4) \u003d H 3 PO 4; M (1 H 3 PO 4) \u003d 98 g / mol.

In reaction (2), phosphoric acid exchanges three atoms f e (H 3 PO 4) hydrogen for the metal, i.e. behaves like a dibasic acid, therefore f e (H 3 PO 4) in reaction (2) is equal to 1/2;

E (H 3 PO 4) \u003d 1 / 2H 3 PO 4; M (1/2 H 3 PO 4) \u003d 49 g / mol.

In reaction (3), phosphoric acid behaves like a tribasic acid, therefore f e (H 3 PO 4) in this reaction is equal to 1/3;

E (H 3 PO 4) \u003d 1 / 3H 3 PO 4; M (1/3 H 3 PO 4) \u003d 32.67 g / mol.

Task 3.2.2 An excess of potassium hydroxide affected the solutions: a) potassium dihydrogen phosphate; b) dihydroxovismuth (III) nitrate. Write the reaction equations for these SKON substances and determine their equivalent and molar masses of equivalents.

Solution: Let's write down the equations of the occurring reactions:

KN 2 PO 4 + 2KON \u003d K 3 PO 4 + 2 H 2 O

Bi (OH) 2 NO 3 + KOH \u003d Bi (OH) 3 + KNO 3

Potassium dihydrogen phosphate, when interacting with KOH, exchanges two hydrogen atoms for a metal, i.e. f e salt KN 2 PO 4 \u003d 1/2;

E (KN 2 PO 4) \u003d 1/2 KN 2 PO 4;

M (1/2 KN 2 PO 4) \u003d 1/2 (39 + 2 + 31 +64) \u003d 68g / mol

Dihydroxovismuth (III) nitrate in the reaction with SCON exchanges one group NO 3 - for ionOH -, i.e. in the basic salt molecule, one bond is rearranged, therefore:

E (Bi (OH) 2 NO 3) \u003d Bi (OH) 2 NO 3;

M (1 Bi (OH) 2 NO 3) \u003d 305 g / mol

Equilibrium theme

Usually problems on this topic turn out to be difficult, because specific knowledge of chemistry alone does not help to solve them; but the applicant is required to have a “mathematical vision” of the problem and to translate chemical quantities (mol) into fairly simple algebraic expressions. Not everyone is ready for the fact that on the chemistry exam they will have to think "mathematically" rather than remembering what they have learned.

The problems given below are taken from the collection: Kuzmenko N.E., Eremin V.V., Churanov S.S., Collection of competitive problems in chemistry - M .: Exam, 2001 - 576 p. ; in [square brackets] are the pages where conditions and solutions (answers) are located.

Tasks

1. (Khimfak-97, option PO-97-1 [p. 290])
One mole of ammonia was placed in a 20 L vessel and heated to 600 0 C. The pressure in the vessel was found to be 435 kPa. Calculate the decomposition rate of ammonia.

2. (Khimfak-spring-98; VKNM-98; Khimfak-correspondence course-99, version SO-98-1 [p. 93])
Three moles of substances A, B, C were mixed. After equilibrium was established, A + B \u003d 2C, 5 moles of substance C were found in the system. Calculate the equilibrium constant. Determine the equilibrium composition of the mixture (in mol%) obtained by mixing substances A, B, C in a molar ratio of 3: 2: 1 at the same temperature.

3. (Faculty of Chemistry-Spring-93; Faculty of Chemistry-Correspondence-94; Chemistry Faculty-Spring-94; VKNM-96, version 171-94-2 [p. 55]. This is one of the most difficult competitive tasks of the Chemistry Department of Moscow State University)
There is a mixture of nitrogen and hydrogen, which is 5% lighter than helium. After passing the mixture over the heated catalyst, ammonia was formed, as a result of which the mixture became heavier than helium under the same conditions. Calculate the range of acceptable values \u200b\u200bfor the reaction yield.

Equilibrium theme

1 ... [Collection, p. 560]
The amount (mol) of gases after the reaction: PV / RT \u003d 435 * 20 / (8.31 * 873) \u003d 1.20 mol
If x mol of ammonia has decomposed, then the decomposition scheme: NH 3 (1-x) N 2 (x / 2) + H 2 (3x / 2)
From the equation: 1.20 mol \u003d (1-x) + x / 2 + 3x / 2 \u003d 1 + x
we get x \u003d 0.2 mol.
Answer: Decomposition of ammonia 20%

2 ... [Collection, p. 412]

K \u003d (1 + 2x) 2 / ((3-x) (2-x)) \u003d 6.25
x \u003d 1.115

Answer: Mole fractions of substances in an equilibrium mixture:
(A) \u003d (3-1,115) / 6 \u003d 0,314 ;
(B) \u003d (2-1,115) / 6 \u003d 0,148 ;
(C) \u003d 0,538

3 ... [Collection, p. 371]
Let the initial mixture contain X mol of N 2 and Y mol of H 2.
The average molar mass of the mixture is 5% lighter than helium or 0.95 * 4:
M Wed \u003d (28X + 2Y) / (X + Y) \u003d 0.95 * 4 \u003d 3.8;
Then Y \u003d 13.44X
Reaction: N 2 + 3 H 2 \u003d 2 NH 3
If a mol of N 2 and 3a mol of H 2 have reacted, we get after the reaction:
(X - a) + (Y - 3a) + 2a \u003d 14.44X - 2a (mol)
The mass of the mixture after the reaction (express through X, since Y \u003d 13.44X):
28X + 2Y \u003d 54.9 X g
Average molar mass of the mixture after the reaction\u003e 4 g / mol (as required):
M Wed \u003d 54.9X / (14.44X - 2a)\u003e 4;
then: a\u003e 0.3575X
The reaction yield is the fraction of reacted nitrogen (coefficient in reaction 1): a / X;
The mixture will become heavier than helium (M av.\u003e 4) at a / X\u003e 35.75%
Answer: the output of ammonia is more than 35.75%

The topic "Equilibrium in solutions"
Tasks

The topic "equilibrium in solutions" is considered difficult because it uses concepts that are only included in the curriculum for schools and classes with advanced study of chemistry - the product of solubility and pH. But the main difficulty is not in the fairly simple formulas themselves, but in the ability to use them in a wide range of problem conditions.

The problems of 2002 are taken from the collection of problems of the past academic year, published annually at the Faculty of Chemistry: "MSU-2002 Written Exam in Chemistry" Chem. Faculty of Moscow State University, 2002.

Tasks

1) (Chemical Faculty, 2002) 500 ml of a saturated solution of Zn 3 (PO 4) 2 contains 2.47 * 10 –7 mol of phosphate ions. Calculate the solubility of the salt in mol / L and the solubility product of Zn 3 (PO 4) 2
2) (VKNM-96, variant YuM-96-1, [p. 240]) Determine the molar concentration of a saturated solution of iron (II) hydroxide at 25 0 С if its solubility product at this temperature is 1 * 10 –15
3) (Chem. F-t, 1993, option 171-93-4 [p. 49]) Acetic acid weighing 25 g is dissolved in water, and the volume of the solution is adjusted to 1 liter. Determine the concentration of H + ions in the resulting solution if the dissociation constant of acetic acid is 1.8 * 10 –5. Neglect the change in acid concentration during dissociation.
4) (F-t Fundam. Medicine - 2002) The dissociation constant of acetic acid is 1.75 * 10 –5. Calculate: a) the pH of a 0.1 M solution of this acid; b) pH of a solution containing 0.1 mol / L of this acid and 0.1 mol / L of sodium acetate

Solutions:

1) (collection 2002, p. 44)
1 liter contains phosphate ions: 2.47 * 10 –7 (1000/500) \u003d 4.94 * 10 –7 mol / l. The solution will contain 2 times less than phosphate ions, formula units of zinc phosphate Zn 3 (PO 4) 2: 4.94 * 10 –7 / 2 \u003d 2.47 * 10 -7 mol / l
The solubility product is defined as a constant of heterogeneous equilibrium under the assumption that a poorly soluble substance passes into solution only in the form of ions.
Then for the process:

taking for from molar concentration of iron hydroxide, we get:

(s) (2s) 2 \u003d 4s 3 \u003d 1 * 10 -15
Then c \u003d (0.25. 10 –15) 1/3 \u003d (250. 10 –18) 1/3 \u003d 6.3. 10 -6 (mol / l)
Answer: c (Fe (OH) 2) \u003d 6.3 * 10 -6 mol / l

3. [Collection, p. 361]
Acetic acid is weak, and the concentration of H + ions in its solution is not equal to the concentration of the acid, as in the case of dilute solutions of strong acids.
The dissociation of acetic acid can be simplified as an equilibrium: CH 3 COOH H + + CH 3 COO -
Equilibrium constant, aka dissociation constant:
K d \u003d () /
1 liter contains 25/60 \u003d 0.417 mol to-you; denote the degree of its dissociation, equal to the ratio of the dissociated molecules to the total number of molecules in the solution. The concentration of H + ions (mol / L) is determined from the concentration of the acid and the degree of its dissociation: \u003d fromSince the value is unknown to us, it must be expressed in terms of known values \u200b\u200b- the concentration of acid c and its dissociation constant K d.
If the acid concentration c, then dissociation will give from mole of H + ions and the same number of CH 3 COO - ions. The solution will remain (1-) from mol CH 3 COOH.
Then the dissociation constant:

With a small degree of dissociation (<< 1) можно приближенно считать, что (1-)from equals from... Then K d 2 s; (K d / s) 1/2:
(K d / s) 1/2 \u003d 6.6. 10 –3; \u003d c \u003d 6.6. 10 –3. 0.417 \u003d 2.74 * 10 -3 mol / l
Answer: 2.74 * 10 -3 mol / l

4) (collection 2002, p. 59)
This is a typical buffer solution problem. But it is unlikely that applicants know (and do not have to know) ready-made formulas for calculating the pH of buffer solutions - this topic is not in the school curriculum or in the program for applicants to Moscow State University. Therefore, for calculations, only the known expressions for the equilibrium constant of a weak acid, the value of the ionic product of water and the determination of pH should be used: pH \u003d - lg, where the entry in square brackets means that the concentrations are expressed in mol / L.

a) Acetic acid dissociation constant:

K d \u003d /
Since \u003d, you can write: 2 \u003d K d. Since acetic acid is a weak electrolyte with a small dissociation constant, one can neglect the fact that part of the original acid dissociated and equate the acid concentration in the expression for the equilibrium constant to the initial (total) concentration: С CH3COOH.

Then we get: 2 \u003d K d C CH3COOH;
\u003d (K d C CH3COOH) 1/2 \u003d (1.75. 10 –5. 10 –1) 1/2 \u003d 1.32. 10 –3; pH \u003d - lg \u003d 2.88

b) Sodium acetate (salt), unlike acetic acid, dissociates completely. Therefore, in the formula for the dissociation constant describing equilibrium, we obtain: K d \u003d /;
With CH3COOH; \u003d C CH3COONa \u003d 0.1 mol / l.
Then: \u003d K d. C CH3COOH / C CH3COONa \u003d 1.75. 10 -5. 10 -1 / 10 -1 \u003d 1.75. 10 –5;
pH \u003d 4.76
Answer: a) pH \u003d 2.88; b) pH \u003d 4.76

Dahl's Explanatory Dictionary

Female aphid (from small) tiny twilight (butterfly), dagger; his caterpillar, which sharpens furs and woolen clothes, Tinca. There are moths for fur coat, for clothes, for cheese, for bread, for vegetables. Moth disappears from hops, camphor. | Vegetable moth, aphid, moth, ... ... Dahl's Explanatory Dictionary

1. MOLE, and; g. A small butterfly, whose caterpillar is a pest of woolen things, grain grains and plants. 2. MOLE, and; f .; MOLE, I; m. Special. Timber floated down the river with logs not tied into a raft. The m was floating along the river. Making my way by boat ... ... encyclopedic Dictionary

MOLE - a unit of the amount of a substance in SI, defined as the amount of a substance containing the same number of formula (structural) units of this substance (atoms, molecules, ions, electrons, etc.) as there are atoms in 12 g of the isotope of carbon 12 (12C); ... ... Big Polytechnic Encyclopedia

Source by moth .. Dictionary of Russian synonyms and expressions similar in meaning. under. ed. N. Abramova, M .: Russian dictionaries, 1999. moth aphids, moths Dictionary of Russian synonyms ... Synonym dictionary

1) the name of the beer in Nimwegen. 2) woolen fabric. 3) for beekeepers: the plexus at the top of the hive. Dictionary of foreign words included in the Russian language. Chudinov A.N., 1910. mol 1 it. molle soft) muses. the same as minor 1 (opposed to fools). 2 ... ... Dictionary of foreign words of the Russian language

mole - Unit number of things, i.e. magnitudes, estimating. number of contents in physical. identity system. structures, elements (atoms, molecules, ions and other particles or their specific. groups), m is equal to the number of things in the system, containing. the same number of structural elements ... ... Technical translator's guide

MOL (Mohl) Hugo von (1805 1872), German botanist, pioneer in the study of the anatomy and physiology of plant CELLS. He formulated a hypothesis that the cell nucleus is surrounded by a granular colloidal substance, which in 1846 he named ... ... Scientific and technical encyclopedic dictionary

MOLE, and, wives. A small butterfly, a caterpillar to the swarm is a pest of fur, wool, grain grains, plants. Moth-eaten (also translated: about which n. Clearly outdated, obsolete; ne.). | adj. Mole, oh, oh. II. MOLE, me, husband. (specialist.). Wood rafting ... ... Ozhegov's Explanatory Dictionary

- (mol, mol), units SI count in va. 1 M. contains as many molecules (atoms, ions, or other structural elements in VA) as there are atoms in 0.012 kg of 12C (a carbon nuclide with an atomic mass of 12) (see AVOGADRO CONSTANT). Physical ... ... Physical encyclopedia

MOLE, meaning unchanged adj. (music). The same as molar. Sonata a Mole. Ushakov's explanatory dictionary. D.N. Ushakov. 1935 1940 ... Ushakov's Explanatory Dictionary

Books

• Mole for Mr. L Cupid, Lydia Scriabin. This is a psychological novel about love, money and love of money. About how the previously forbidden and condemned painfully sweet ideas of "personal enrichment" burst into modern life ...

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

Mole – this is the 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 the carbon isotope 12 FROM .

Considering 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 AND 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 in the amount of one mole is 6.02 10 23 and called avogadro number (N AND ).

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 designated by the letter M and has the dimension [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:

Мr (Н 2 О) \u003d 2Аr (Н) + Аr (O) \u003d 2 ∙ 1 + 16 \u003d 18 amu.

The molar mass of water is the same, but 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. In water, the amount of substance is 1 mol, contains 2 mol of hydrogen atoms and one mol of oxygen atoms.

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

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

where ν is the amount of substance, [mol]; m - mass of 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 \u200b\u200bof the relative atomic masses:

Мr (Na 2 SO 4) \u003d 2Аr (Na) + Аr (S) + 4Аr (O) \u003d 142,

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

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

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

m \u003d ν M = 5 mol 142 g / mol \u003d 710 g.

Answer: 710.

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

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 different gases and vapors occupies the same volume, equal to 22.4 l.

The volume occupied by 1 mole of gas or vapor at normal conditions is called molar volume gas and has a dimension of 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 \u003d 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.

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

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1 mol per liter [mol / l] \u003d 1000 mol per meter³ [mol / m³]

Initial value

Converted value

mol per meter³ mol per liter mol per centimeter³ mol per millimeter³ kilomole per meter³ kilomole per liter kilomole per centimeter³ kilomole per millimeter³ millimole per meter³ millimol per liter millimole per centimeter³ millimole per millimeter³ mol per cubic meter. decimeter molar millimolar micromolar nanomolar picomolar femtomolar attomolar zeptomolar ioctomolar

More on molar concentration

General information

The concentration of a solution can be measured in various ways, for example, 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 in a small amount of a substance there is usually a large number of elementary components, special units, moles, are used to measure the amount of a substance. One mole equals the number of atoms in 12 grams of carbon-12, which is approximately 6 × 10²³ atoms.

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

Molarity should not be confused with molality... Unlike molarity, molality is the ratio of the amount of soluble substance to the mass of the solvent, not to the mass of the entire solution. When the solvent is water, and the amount of soluble substance is small compared to the amount of water, then molarity and molality are similar in meaning, but in other cases 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, as the temperature rises, the solvent evaporates, and the amount of the soluble substance does not change - in this case, the concentration of the solution will increase. Sometimes the opposite happens. Sometimes a change in temperature affects how the soluble substance dissolves. For example, some or all of the soluble substance stops dissolving, and the concentration of the solution decreases.

Units

Molar concentration is measured in moles per unit volume, for example moles per liter or moles per cubic meter. Moles per cubic meter is the 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 the substance. The amount of a substance can be calculated using the chemical formula of this substance and information about the total mass of this substance in solution. That is, to find out the amount of solution in moles, we learn from the periodic table the atomic mass of each atom in the solution, and then divide the total mass of the substance by the total atomic mass of atoms in the molecule. Before adding the atomic mass together, make sure we multiply the mass of each atom by the number of atoms in the molecule we are looking at.

Calculations can be performed in reverse order. If you know the molar concentration of the solution and the formula of the soluble substance, then you can find out the amount of solvent in the solution, in moles and grams.

Examples of

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

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

Since oxygen in the formula is O₃, it is necessary to multiply the atomic mass of oxygen by 3. We get 47.9982. Now we 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. Our task is 51 grams of soda. Find the molar mass by dividing 51 grams by the mass of one mole, that is, 84 grams, and we get 0.6 mol.

It turns out that our solution is 0.6 mole 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 soluble substance were used in the solution, its concentration is low.

Let's look at another example. Find the molar concentration of one sugar cube in a cup of tea. Table sugar is composed of sucrose. First, we 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 \u003d 342 grams. There are 4 grams of sugar in one cube, which gives us 4/342 \u003d 0.01 moles. There are about 237 milliliters of tea in one cup, which means that the concentration of sugar in one cup of tea is 0.01 mol / 237 milliliters × 1000 (to convert milliliters to liters) \u003d 0.049 mol per liter.

Application

Molar concentration is widely used in calculations related to chemical reactions. The section of chemistry, in which the ratios between substances in chemical reactions are calculated and often work with moles, is called stoichiometry... The molar concentration can be found by the chemical formula of the final product, which then becomes a soluble substance, as in the example with a soda solution, but you can also first find this substance by 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 equation of the chemical reaction, we find out the formula of the molecule of the dissolved substance, and then we find the mass of the molecule and the molar concentration using the periodic table, as in the examples above. Of course, calculations can be made in reverse order using information on the molar concentration of a substance.

Let's look at a simple example. This time we'll mix baking soda and vinegar to see an interesting chemical reaction. Both vinegar and soda are easy to find - you probably have them in your kitchen. As mentioned above, the formula for 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, since the concentration of vinegar is different in different countries. In this experiment, you don't 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 later we will calculate the concentration of the solution.

First, let's solve the equation for the chemical reaction between soda and acetic acid:

NaHCO₃ + CH₃COOH → NaC₂H₃O₂ + H₂CO₃

The reaction product is H₂CO₃, a substance that reacts chemically again due to its low stability.

H₂CO₃ → H₂O + CO₂

The reaction produces water (H получаемO), carbon dioxide (CO₂) and sodium acetate (NaC₂H₃O₂). We will mix the obtained 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 the acetic acid is dissolved. Sodium acetate is an interesting substance. It is used in chemical warmers 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 find the molar concentration later, it should be noted that only a limited amount of a substance can react with other substances. It 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 the starting products by the reverse calculation. This method is often used in practice, in calculations related to chemical reactions.

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

In pharmaceuticals

When creating drugs, molar concentration is very important, since it determines how the drug affects the body. If the concentration is too high, the drugs can even be lethal. On the other hand, if the concentration is too low, then the medicine 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 with its help is found osmotic concentration... Osmotic concentration is used more often than molar concentration. If the concentration of a substance, for example a drug, is higher on one side of the membrane, compared to the concentration on the other side of the membrane, for example, inside the eye, then the more concentrated solution will move through the membrane to where the concentration is lower. This flow of solution through the membrane is often problematic. For example, if fluid moves into a cell, such as a blood cell, it is possible that this fluid overflow will damage the membrane and rupture. Leakage of fluid from the cell is also problematic, as this will disrupt the cell's performance. Any drug-induced flow of fluid through the membrane from the cell or into the cell is desirable to prevent, and for this, the concentration of the drug is tried to be similar to the concentration of fluid in the body, for example, in the blood.

It should be noted that in some cases the molar and osmotic concentration are equal, but this is not always the case. It depends on whether the substance dissolved in water has disintegrated into ions in the process electrolytic dissociation... When calculating the osmotic concentration, particles in general are taken into account, while when calculating the molar concentration, only certain particles, such as molecules, are taken into account. Therefore, if, for example, we work with molecules, but the substance has disintegrated 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, you need to know the physical properties of the solution.

In the manufacture of medicines, pharmacists also take into account tonicity solution. Tonality is a property of a solution that depends on concentration. Unlike osmotic concentration, tonicity is the concentration of substances that the membrane does not pass. The osmosis process forces solutions with a higher concentration to move into solutions with a lower concentration, but if the membrane prevents this movement by not allowing the solution to pass through it, then there is pressure on the membrane. Such pressure is usually problematic. If a drug is intended to penetrate the blood or other fluid in the body, then it is necessary to balance the tonicity of this drug with the tonicity of the body fluid in order 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 with a concentration that allows you to balance the tonicity of the body fluid and the tonicity of the mixture of this solution and medicine. Usually, the isotonic solution is stored in sterile containers and infused intravenously. Sometimes it is used in pure form, and sometimes as a mixture with a medicine.

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