Introduction to General Chemistry. Questions for self-control In 1000 g of water
The relative masses of atoms and molecules are determined using the D.I. Mendeleev values of atomic masses. At the same time, when carrying out calculations for educational purposes, the values of the atomic masses of the elements are usually rounded to integers (with the exception of chlorine, whose atomic mass is assumed to be 35.5).
Example 1 Relative atomic mass of calcium And r (Ca)=40; relative atomic mass of platinum And r (Pt)=195.
The relative mass of a molecule is calculated as the sum of the relative atomic masses of the atoms that make up this molecule, taking into account the amount of their substance.
Example 2 Relative molar mass sulfuric acid:
M r (H 2 SO 4) \u003d 2A r (H) + A r (S) + 4A r (O) \u003d 2 · 1 + 32 + 4· 16 = 98.
The absolute masses of atoms and molecules are found by dividing the mass of 1 mole of a substance by the Avogadro number.
Example 3. Determine the mass of one atom of calcium.
Solution. The atomic mass of calcium is And r (Ca)=40 g/mol. The mass of one calcium atom will be equal to:
m (Ca) \u003d A r (Ca) : N A \u003d 40: 6.02 · 10 23 = 6,64· 10 -23 years
Example 4 Determine the mass of one molecule of sulfuric acid.
Solution. The molar mass of sulfuric acid is M r (H 2 SO 4) = 98. The mass of one molecule m (H 2 SO 4) is:
m (H 2 SO 4) \u003d M r (H 2 SO 4) : N A \u003d 98: 6.02 · 10 23 = 16,28· 10 -23 years
2.10.2. Calculation of the amount of matter and calculation of the number of atomic and molecular particles from known values of mass and volume
The amount of a substance is determined by dividing its mass, expressed in grams, by its atomic (molar) mass. The amount of a substance in the gaseous state at n.o. is found by dividing its volume by the volume of 1 mol of gas (22.4 l).
Example 5 Determine the amount of sodium substance n(Na) in 57.5 g of metallic sodium.
Solution. The relative atomic mass of sodium is And r (Na)=23. The amount of a substance is found by dividing the mass of metallic sodium by its atomic mass:
n(Na)=57.5:23=2.5 mol.
Example 6 . Determine the amount of nitrogen substance, if its volume at n.o. is 5.6 liters.
Solution. The amount of nitrogen substance n(N 2) we find by dividing its volume by the volume of 1 mol of gas (22.4 l):
n(N 2) \u003d 5.6: 22.4 \u003d 0.25 mol.
The number of atoms and molecules in a substance is determined by multiplying the number of atoms and molecules in the substance by Avogadro's number.
Example 7. Determine the number of molecules contained in 1 kg of water.
Solution. The amount of water substance is found by dividing its mass (1000 g) by the molar mass (18 g / mol):
n (H 2 O) \u003d 1000: 18 \u003d 55.5 mol.
The number of molecules in 1000 g of water will be:
N (H 2 O) \u003d 55.5 · 6,02· 10 23 = 3,34· 10 24 .
Example 8. Determine the number of atoms contained in 1 liter (n.o.) of oxygen.
Solution. The amount of oxygen substance, the volume of which under normal conditions is 1 liter is equal to:
n(O 2) \u003d 1: 22.4 \u003d 4.46 · 10 -2 mol.
The number of oxygen molecules in 1 liter (N.O.) will be:
N (O 2) \u003d 4.46 · 10 -2 · 6,02· 10 23 = 2,69· 10 22 .
It should be noted that 26.9 · 10 22 molecules will be contained in 1 liter of any gas at n.o. Since the oxygen molecule is diatomic, the number of oxygen atoms in 1 liter will be 2 times greater, i.e. 5.38 · 10 22 .
2.10.3. Calculation of the average molar mass of the gas mixture and volume fraction
the gases it contains
The average molar mass of a gas mixture is calculated from the molar masses of the constituent gases of this mixture and their volume fractions.
Example 9 Assuming that the content (in volume percent) of nitrogen, oxygen and argon in the air is 78, 21 and 1, respectively, calculate the average molar mass of air.
Solution.
M air = 0.78 · M r (N 2)+0.21 · M r (O 2)+0.01 · M r (Ar)= 0.78 · 28+0,21· 32+0,01· 40 = 21,84+6,72+0,40=28,96
Or approximately 29 g/mol.
Example 10 gas mixture contains 12 l NH 3 , 5 l N 2 and 3 l H 2 measured at n.o. Calculate the volume fractions of gases in this mixture and its average molar mass.
Solution. The total volume of the mixture of gases is V=12+5+3=20 l. Volume fractions j of gases will be equal:
φ(NH 3)= 12:20=0.6; φ(N 2)=5:20=0.25; φ(H 2)=3:20=0.15.
The average molar mass is calculated on the basis of the volume fractions of the constituent gases of this mixture and their molecular masses:
M=0.6 · M (NH 3) + 0.25 · M(N2)+0.15 · M (H 2) \u003d 0.6 · 17+0,25· 28+0,15· 2 = 17,5.
2.10.4. Calculation of the mass fraction of a chemical element in a chemical compound
The mass fraction ω of a chemical element is defined as the ratio of the mass of an atom of a given element X contained in a given mass of a substance to the mass of this substance m. Mass fraction is a dimensionless quantity. It is expressed in fractions of a unit:
ω(X) = m(X)/m (0<ω< 1);
or in percentage
ω(X),%= 100 m(X)/m (0%<ω<100%),
where ω(X) is the mass fraction of the chemical element X; m(X) is the mass of the chemical element X; m is the mass of the substance.
Example 11 Calculate the mass fraction of manganese in manganese (VII) oxide.
Solution. The molar masses of substances are equal: M (Mn) \u003d 55 g / mol, M (O) \u003d 16 g / mol, M (Mn 2 O 7) \u003d 2M (Mn) + 7M (O) \u003d 222 g / mol. Therefore, the mass of Mn 2 O 7 with the amount of substance 1 mol is:
m(Mn 2 O 7) = M(Mn 2 O 7) · n(Mn 2 O 7) = 222 · 1= 222
From the formula Mn 2 O 7 it follows that the amount of substance of manganese atoms is twice the amount of substance of manganese oxide (VII). Means,
n(Mn) \u003d 2n (Mn 2 O 7) \u003d 2 mol,
m(Mn)= n(Mn) · M(Mn) = 2 · 55 = 110 g.
Thus, the mass fraction of manganese in manganese(VII) oxide is:
ω(X)=m(Mn) : m(Mn 2 O 7) = 110:222 = 0.495 or 49.5%.
2.10.5. Establishing the formula of a chemical compound by its elemental composition
The simplest chemical formula of a substance is determined on the basis of the known values of the mass fractions of the elements that make up this substance.
Suppose there is a sample of a substance Na x P y O z with a mass m o g. Consider how its chemical formula is determined if the quantities of the substance of the atoms of the elements, their masses or mass fractions in the known mass of the substance are known. The formula of a substance is determined by the ratio:
x: y: z = N(Na) : N(P) : N(O).
This ratio does not change if each of its terms is divided by Avogadro's number:
x: y: z = N(Na)/N A: N(P)/N A: N(O)/N A = ν(Na) : ν(P) : ν(O).
Thus, to find the formula of a substance, it is necessary to know the ratio between the amounts of substances of atoms in the same mass of a substance:
x: y: z = m(Na)/M r (Na) : m(P)/M r (P) : m(O)/M r (O).
If we divide each term of the last equation by the mass of the sample m o , then we get an expression that allows us to determine the composition of the substance:
x: y: z = ω(Na)/M r (Na) : ω(P)/M r (P) : ω(O)/M r (O).
Example 12. The substance contains 85.71 wt. % carbon and 14.29 wt. % hydrogen. Its molar mass is 28 g/mol. Determine the simplest and true chemical formulas of this substance.
Solution. The ratio between the number of atoms in a C x H y molecule is determined by dividing the mass fractions of each element by its atomic mass:
x: y \u003d 85.71 / 12: 14.29 / 1 \u003d 7.14: 14.29 \u003d 1: 2.
Thus, the simplest formula of a substance is CH 2. The simplest formula of a substance does not always coincide with its true formula. In this case, the formula CH 2 does not correspond to the valency of the hydrogen atom. To find the true chemical formula, you need to know the molar mass of a given substance. In this example, the molar mass of the substance is 28 g/mol. Dividing 28 by 14 (the sum of atomic masses corresponding to the formula unit CH 2), we obtain the true ratio between the number of atoms in a molecule:
We get the true formula of the substance: C 2 H 4 - ethylene.
Instead of the molar mass for gaseous substances and vapors, the density for any gas or air can be indicated in the condition of the problem.
In the case under consideration, the gas density in air is 0.9655. Based on this value, the molar mass of the gas can be found:
M = M air · D air = 29 · 0,9655 = 28.
In this expression, M is the molar mass of gas C x H y, M air is the average molar mass of air, D air is the density of gas C x H y in air. The resulting value of the molar mass is used to determine the true formula of the substance.
The condition of the problem may not indicate the mass fraction of one of the elements. It is found by subtracting from unity (100%) the mass fractions of all other elements.
Example 13 An organic compound contains 38.71 wt. % carbon, 51.61 wt. % oxygen and 9.68 wt. % hydrogen. Determine the true formula of this substance if its oxygen vapor density is 1.9375.
Solution. We calculate the ratio between the number of atoms in the molecule C x H y O z:
x: y: z = 38.71/12: 9.68/1: 51.61/16 = 3.226: 9.68: 3.226= 1:3:1.
The molar mass M of a substance is:
M \u003d M (O 2) · D(O2) = 32 · 1,9375 = 62.
The simplest formula of a substance is CH 3 O. The sum of atomic masses for this formula unit will be 12+3+16=31. Divide 62 by 31 and get the true ratio between the number of atoms in the molecule:
x:y:z = 2:6:2.
Thus, the true formula of the substance is C 2 H 6 O 2. This formula corresponds to the composition of dihydric alcohol - ethylene glycol: CH 2 (OH) -CH 2 (OH).
2.10.6. Determination of the molar mass of a substance
The molar mass of a substance can be determined on the basis of its gas vapor density with a known molar mass.
Example 14 . The vapor density of some organic compound in terms of oxygen is 1.8125. Determine the molar mass of this compound.
Solution. The molar mass of an unknown substance M x is equal to the product of the relative density of this substance D by the molar mass of the substance M, according to which the value of the relative density is determined:
M x = D · M = 1.8125 · 32 = 58,0.
Substances with the found value of the molar mass can be acetone, propionaldehyde and allyl alcohol.
The molar mass of a gas can be calculated using the value of its molar volume at n.c.
Example 15. Mass of 5.6 liters of gas at n.o. is 5.046 g. Calculate the molar mass of this gas.
Solution. The molar volume of gas at n.s. is 22.4 liters. Therefore, the molar mass of the desired gas is
M = 5.046 · 22,4/5,6 = 20,18.
The desired gas is neon Ne.
The Clapeyron–Mendeleev equation is used to calculate the molar mass of a gas whose volume is given under non-normal conditions.
Example 16 At a temperature of 40 ° C and a pressure of 200 kPa, the mass of 3.0 liters of gas is 6.0 g. Determine the molar mass of this gas.
Solution. Substituting the known quantities into the Clapeyron–Mendeleev equation, we obtain:
M = mRT/PV = 6.0 · 8,31· 313/(200· 3,0)= 26,0.
The gas under consideration is acetylene C 2 H 2.
Example 17 Combustion of 5.6 l (N.O.) of hydrocarbon produced 44.0 g of carbon dioxide and 22.5 g of water. The relative density of the hydrocarbon with respect to oxygen is 1.8125. Determine the true chemical formula of the hydrocarbon.
Solution. The reaction equation for the combustion of hydrocarbons can be represented as follows:
C x H y + 0.5 (2x + 0.5y) O 2 \u003d x CO 2 + 0.5 y H 2 O.
The amount of hydrocarbon is 5.6:22.4=0.25 mol. As a result of the reaction, 1 mol of carbon dioxide and 1.25 mol of water are formed, which contains 2.5 mol of hydrogen atoms. When a hydrocarbon is burned with a quantity of a substance of 1 mole, 4 moles of carbon dioxide and 5 moles of water are obtained. Thus, 1 mol of hydrocarbon contains 4 mol of carbon atoms and 10 mol of hydrogen atoms, i.e. chemical formula of hydrocarbon C 4 H 10 . The molar mass of this hydrocarbon is M=4 · 12+10=58. Its relative oxygen density D=58:32=1.8125 corresponds to the value given in the condition of the problem, which confirms the correctness of the found chemical formula.
Problem 427.
Calculate the mole fractions of alcohol and water in a 96% (by mass) solution of ethyl alcohol.
Solution:
Mole fraction(N i) - the ratio of the amount of a solute (or solvent) to the sum of the amounts of all
substances in solution. In a system consisting of alcohol and water, the mole fraction of water (N 1) is equal to
And the mole fraction of alcohol 
, where n 1 - the amount of alcohol; n 2 - the amount of water.
We calculate the mass of alcohol and water contained in 1 liter of solution, provided that their densities are equal to one from the proportions:
a) mass of alcohol:
b) mass of water:
We find the amount of substances according to the formula: , where m (B) and M (B) - the mass and amount of the substance.
Now we calculate the mole fractions of substances:
Answer: 0,904; 0,096.
Problem 428.
666 g of KOH are dissolved in 1 kg of water; the density of the solution is 1.395 g/ml. Find: a) mass fraction of KOH; b) molarity; c) molality; d) mole fractions of alkali and water.
Solution:
but) Mass fraction- the percentage of the mass of the dissolved substance to the total mass of the solution is determined by the formula:
where
m (solution) \u003d m (H 2 O) + m (KOH) \u003d 1000 + 666 \u003d 1666
b) Molar (volume-molar) concentration shows the number of moles of a solute contained in 1 liter of solution.
Let's find the mass of KOH per 100 ml of solution according to the formula: formula: m = p V, where p is the density of the solution, V is the volume of the solution.
m(KOH) = 1.395 . 1000 = 1395
Now we calculate the molarity of the solution:
We find how many grams of HNO 3 are per 1000 g of water, making up the proportion:
d) Mole fraction (N i) - the ratio of the amount of a dissolved substance (or solvent) to the sum of the amounts of all substances in solution. In a system consisting of alcohol and water, the mole fraction of water (N 1) is equal to and the mole fraction of alcohol, where n 1 is the amount of alkali; n 2 - the amount of water.
100g of this solution contains 40g KOH 60g H2O.
Answer: a) 40%; b) 9.95 mol/l; c) 11.88 mol/kg; d) 0.176; 0.824.
Problem 429.
The density of a 15% (by weight) solution of H 2 SO 4 is 1.105 g/ml. Calculate: a) normality; b) molarity; c) the molality of the solution.
Solution:
Let's find the mass of the solution using the formula: m = p V, where p is the density of the solution, V is the volume of the solution.
m(H 2 SO 4) = 1.105 . 1000 = 1105
The mass of H 2 SO 4 contained in 1000 ml of solution is found from the proportion:
Let's determine the molar mass of the equivalent of H 2 SO 4 from the ratio:
M E (B) - molar mass of the acid equivalent, g / mol; M(B) is the molar mass of the acid; Z(B) - equivalent number; Z(acids) is equal to the number of H+ ions in H 2 SO 4 → 2.
a) Molar equivalent concentration (or normality) indicates the number of equivalents of a solute contained in 1 liter of solution.
b) Molar concentration
Now we calculate the molality of the solution:
c) Molar concentration (or molality) shows the number of moles of a solute contained in 1000 g of solvent.
We find how many grams of H 2 SO 4 are contained in 1000 g of water, making up the proportion:
Now we calculate the molality of the solution:
Answer: a) 3.38n; b) 1.69 mol/l; 1.80 mol/kg.
Problem 430.
The density of a 9% (by weight) solution of sucrose C 12 H 22 O 11 is 1.035 g/ml. Calculate: a) the concentration of sucrose in g/l; b) molarity; c) the molality of the solution.
Solution:
M (C 12 H 22 O 11) \u003d 342 g / mol. Let's find the mass of the solution using the formula: m = p V, where p is the density of the solution, V is the volume of the solution.
m (C 12 H 22 O 11) \u003d 1.035. 1000 = 1035
a) The mass of C 12 H 22 O 11 contained in the solution is calculated by the formula:
where
- mass fraction of the dissolved substance; m (in-va) - the mass of the dissolved substance; m (r-ra) - the mass of the solution.
The concentration of a substance in g / l shows the number of grams (mass units) contained in 1 liter of solution. Therefore, the concentration of sucrose is 93.15 g/l.
b) Molar (volume-molar) concentration (C M) shows the number of moles of a solute contained in 1 liter of solution.
in) Molar concentration(or molality) indicates the number of moles of a solute contained in 1000 g of solvent.
We find how many grams of C 12 H 22 O 11 are contained in 1000 g of water, making up the proportion:
Now we calculate the molality of the solution:
Answer: a) 93.15 g/l; b) 0.27 mol/l; c) 0.29 mol/kg.
What are solutions and what features of chemical compounds and mechanical mixtures do they have?
What determines the thermal effect of dissolution?
What is solubility and what does it depend on?
What is the concentration of a solution? Give the definition of percentage, molar, molar equivalent concentration and molar concentration, as well as mole fraction.
Define Raoult's law.
What are the consequences of Raoult's law?
What are cryoscopic and ebullioscopic solvent constants?
Literature.
Korovin N.V. General chemistry.- M.: Higher. school, 2002. Ch. 8, § 8.1.
Glinka N.L. General Chemistry.- M.: Integral-Press, 2002, Ch. 7,
1.6. Examples of problem solving
Example 1. When dissolving 10 g of potassium nitrate (KNO 3) in 240 g of water, the temperature of the solution decreased by 3.4 degrees. Determine the heat of solution of salt. The specific heat capacity (ssp) of the solution is 4.18 J/g. TO.
Solution:
1. Find the mass of the resulting solution (m):
m = 10 + 240 = 250 (g).
2. Let's determine the amount of heat absorbed by the solution:
Q=m. court. T
Q=250. 4.18. (-3.4) \u003d - 3556.4 J \u003d - 3.56 kJ.
3. We calculate the amount of heat absorbed by dissolving one mole of KNO 3, i.e. its heat of dissolution (the molar mass of KNO 3 is 101 g/mol):
when 10 g of salt is dissolved, 3.56 kJ is absorbed
when dissolving 101 g of salt --------- x,
x = = 35.96 kJ
Answer: the heat of dissolution of KNO 3 is 35.96 kJ/mol.
Solution:
1. We find the weight amount of sulfuric acid contained in 1 liter of 17.5% solution:
a) find the mass of a liter (1000 ml) of the solution:
m = . V = 1.12 . 1000 = 1120 g;
b) find the weight of sulfuric acid:
100 g of solution contains 17.5 g of H 2 SO 4;
in 1120 g of solution - x,
2. We find the titer of the solution; to do this, it is necessary to divide the weight of the acid contained in a known volume of solution by the volume of the solution, expressed in milliliters:
T = = 0.196 g/ml.
3. Find the molar concentration of the solution; for this, it is necessary to divide the weight amount of acid contained in 1 liter of solution by the molar mass (MH 2 SO 4), 98 g / mol:
2 mol/l.
4. Find the molar concentration of the solution equivalent; for this, it is necessary to divide the weight amount of acid contained in 1 liter of solution (196 g) by the equivalent mass (EH 2 SO 4).
The equivalent mass of H 2 SO 4 is equal to its molar mass divided by the number of hydrogen atoms:
Therefore, C eq = = 4 moleq / l.
The molar concentration of the equivalent can also be calculated using the formula
.
5. We calculate the molality of the solution; for this it is necessary to find the number of moles of acid contained in 1000 g of solvent (water).
From previous calculations (see paragraph 3), it is known that 1120 g (1 l) of a solution contains 196 g or 2 moles of H 2 SO 4, therefore, water in such a solution:
1120 - 196 = 924
We make a proportion:
924 g of water accounts for 2 moles of H 2 SO 4
per 1000 g of water - x.
C m \u003d x \u003d \u003d 2.16 mol / 1000 g of water.
Answer: T = 0.196 g/ml; = 2 mol/l; C eq = 4 moleq/l;
With m = 2.16 mol / 1000 g of water.
Example 3 How many milliliters of a 96% solution of H 2 SO 4 ( \u003d 1.84 g / cm 3) will be required to prepare 1 liter of its solution with a molar equivalent concentration of 0.5?
Solution.
1. We calculate the weight amount of H 2 SO 4 required to prepare 1 liter of a solution with a molar equivalent concentration of 0.5 (the equivalent of sulfuric acid is 49 g):
1000 ml of a 0.5 N solution contains 49. 0.5 \u003d 24.5 g H 2 SO 4.
2. We determine the weight amount of the initial (96% th) solution containing 24.5 g of H 2 SO 4:
100 g of solution contains 96 g of H 2 SO 4,
in x g of solution - 24.5 g of H 2 SO 4.
x == 25.52 g
3. Find the required volume of the initial solution by dividing the weight amount of the solution by its density ():
V = = 13.87 ml.
Answer: to prepare 1 l of a solution of sulfuric acid with a molar concentration equivalent of 0.5, you need 13.87 ml of a 96% solution of H 2 SO 4.
Example 4 A solution prepared from 2 kg (m) of ethyl alcohol and 8 kg (g) of water was poured into a car radiator. Calculate the freezing point of the solution. The cryoscopic constant of water K to is equal to 1.86.
Solution.
1. We find the decrease in the freezing point of the solution using the corollary from Raoult's law:
t s \u003d K to C m \u003d K to .
The molar mass of C 2 H 5 OH is 46 g / mol, therefore,
T s \u003d 1.86 \u003d 10.1 o C.
2. Find the freezing point of the solution:
T s \u003d 0 - 10.1 \u003d - 10.1 o C.
Answer: the solution freezes at a temperature of -10.1 o C.