Fundamentals of General Chemistry. Terms and Problems in Tests In 2 parts. P.1. Terms and Examples in Tasks. Study guide
.pdf5 . S O L U T I O N S A N D T H E I R C O N C E N T R A T I O N S
5 . 1 . D i c t i o n a r y
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Amount of substance ( , mol) |
Количество вещества ( , моль) |
Concentration |
Концентрация |
Density ( , g/mL) |
Плотность ( , г/мл) |
Diluted |
Разбавленный |
Equivalent |
Эквивалент |
Еquivalence number |
Число эквивалентности |
Hydrate |
Гидрат |
Liter (L) |
Литр (л) |
Mass (g) |
Масса (г) |
Molar concentration |
Молярная концентрация |
(С, mol/L) or molarity (С, M) |
(С, моль/л), или молярность (С, М) |
Molar mass (М, g/mol) |
Молярная масса (М, г/моль) |
Normal concentration or normality |
Нормальная концентрация, или |
(СN, mol/L or abbreviated as N) |
нормальность (Сэкв, моль экв/л |
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или сокращенно н.) |
Percent concentration |
Процентная концентрация |
(ω, % by mass) |
(ω, %) |
Saturated |
Насыщенный |
Solute |
Растворенное вещество |
Solution |
Раствор |
Solvent |
Растворитель |
5 . 2 . W o r k e d e x a m p l e s
Example 1. Calculate the mass of potassium phosphate, mass and volume of water, required to prepare 250 g of a solution that is 8.5 % K3РO4 by mass.
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Answer
To calculate the mass of K3РO4 solute, use the formula:
mass solute
ω = mass solution ·100 % where ω is the percent concentration of a solute
msolute= 100ω · msolution ,
msolution = msolute + mwater.
Then
m(K3PO4) = 0.085 · 250 g = 21.25 g;
mass of water is
m(H2O) = 250 – 21.25 = 228.75 g.
The volume of water is determined by the formula:
V(H2O) = m(H2O) : ρ(H2О),
where ρ(H2О) is the density of water, which is equal to 1 g/mL. Then
V(H2O) = 228.75 g : 1 g/mL = 228.75 mL 228.8 mL.
Example 2. Calculate the volume (mL) of a solution that has a density of 1.07 g/mL and contains 14.5 % hydrochloric acid by mass required to prepare 2 liters of a solution with concentration 0.1 M.
Answer
The molar concentration (molarity) is the number of moles of a substance dissolved in one liter of solution. To determine the number of HCl moles required to prepare of 2 liters of a 0.1 M solution use the formula:
С = νsolute ,
Vsolution
where C is the molar concentration of a solution (mol/L), νsolute is the number of moles of solute, Vsolution is the volume of the solution (L).
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ν(HCl) = Csolution 2 ·Vsolution 2 = 0.1 mol/L 2 L = 0.2 mol.
The molar mass of HCl is 36.5 g/mol. Calculate the mass of HCl required to prepare 2 liters of a solution with concentration 0.1 M:
m(HCl) = ν(HCl) M(HCl) = 0.2 mol 36.5 g/mol =7.3 g.
Then the mass of the initial solution is:
msolution1 = m(HCl) : ω(HCl)/100 = 7.3 g : 14.5/100 = 50.34 g,
and the volume of the initial HCl solution is:
Vsolution1 = msolution1 : ρ = 50.34 g : 1.07 g/mL = 47.05 mL.
Example 3. Calculate the molar mass of the equivalent of barium hydroxide.
Answer
There is a relationship between equivalent molar mass ( Meq) and molar mass (M) of acid, base, salt:
Meq(acid) = |
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Мacid |
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number of protons |
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Meq(base) = |
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number of hydroxide ions |
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Meq(salt) = |
Msalt |
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q ∙ x |
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where q is the metal cation charge, x is the number of cations in the salt formula.
The units of equivalent molar mass are g/mol. Then
Meq(Ba(OH) ) = |
M(Ba(OH)2) |
; M(Ba(OH) ) = 171.35 g/mol; |
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Meq(Ba(OH)2) = 171.35 g/mol = 85.68 g/mol. 2
Example 4. Calculate the molarity and normality of a solution that has a density of 1.218 g/mL and contains 30 % sulphuric acid by mass.
Answer
Let the mass of the H2SO4 solution is 100 g, then its volume will be
Vsolution = msolution : = 100 g : 1.218 g/mL = 82.1 mL = 0.082 L.
Mass of sulfuric acid in 100 g of a solution is:
m(H2SO4) = ω(H2SO4)/100 msolution = 0.3 100 g = 30 g.
The molarity (molar concentration) of H2SO4 is calculated by the formula:
С = (Н2SO4) = m (Н2SO4).
Vsolution
The molar mass of sulfuric acid: M(H2SO4) = 98 g/mol. Then
30 g
С(H2SO4) = 98 g/mol 0,082 L = 3.73mol/L = 3.73 М.
Normality СN is the number of equivalents νeq of a substance dissolved in one liter of solution, mol/L or abbreviated as N. The normality of a solution is calculated by the formula:
СN = |
νeq |
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msolute |
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Vsolution |
Meq V |
solution |
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where msolute is mass of a dissolved substance; Meq is equivalent molar
mass substance (g/mol); Vsolution is the volume of the solution (L). The equivalent molar mass of sulfuric acid is calculated as
Meq = M(H2SO4)/2 = 98 g/mol / 2 = 49 g/mol.
Then
30 g
СN(H2SO4) = 49 g/mol 0.082 L = 7.46 mol/L = 7.46 N.
To calculate the normality CN of this solution use also the formula from table 5.1:
CN = С ∙ z,
where z is equivalence number. Equivalence number is a positive integer equal to the number of equivalents of a substance contained in one mole of this substance
z = M(substance) . Meq(substance)
For sulfuric acid
z = M(H2SO4)/ Meq(H2SO4) = 98 g/mol / 49 g/mol = 2.
Then
CN = 3.73 2 = 7.46 mol/L = 7.46 N.
Table 5.1
The relationship between different concentrations of a substance in a solution
Concentration |
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С, mol/L |
CN , mol/L |
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C = |
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CN = |
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CN = С z |
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5 . 3 . T a s k s f o r s e l f - c o n t r o l
1. The mass (g) of copper(II) sulfate pentahydrate CuSO4 . 5H2O, required to prepare 400 g of a solution that is 2 % CuSO4 by mass is:
А) 8; |
B) 4.5; |
C) 12.5; |
D) 80. |
2. The mass (g) of potassium carbonate |
required to prepare 1 liter of |
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a 0.6 M potassium carbonate solution is: |
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А) 230; |
B) 83; |
C) 8.3; |
D) 64. |
3. 100 mL of water was added to 25 mL of a solution that has a density of 1.78 g/mL and contains 85.16 % H2SO4 by mass. The percentage concentration of sulfuric acid in the resulting solution is:
А) 2.6; |
B) 38; |
C) 8.3; |
D) 26.2. |
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Molarity |
(M) solution |
that is 70 % |
sulfuric acid by mass |
(ρ = 1.615 g/mL) is: |
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А) 0.011; |
B) 4.44; |
C) 11.5; |
D) 1.15. |
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The molar concentration (mol/L) of a solution that has a density of |
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1.22 g/mL and is 20 % NaOH by mass is: |
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А) 0.61; |
B) 0.006; |
C) 6.6; |
D) 6.1. |
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6. |
The molar concentration (mol/L) of a solution that has a density of |
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1.18 g/mL and is 30 % nitric acid by mass is: |
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А) 3.61; |
B) 5.62; |
C) 22.3; |
D) 0.56. |
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7. |
The molar mass of the equivalent of chromium (III) sulfate is: |
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А) 392; |
B) 65.3; |
C) 130.6; |
D) 196. |
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8. |
The molar mass of the equivalent of iron (III) chloride is: |
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А) 27; |
B) 162.5; |
C) 54.1; |
D) 127. |
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9. |
The molar mass of the equivalent of aluminum hydroxide is: |
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А) 234; |
B) 39; |
C) 78; |
D) 26. |
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10. The normality, СN, of a solution that has a density of 1.08 g/mL and is 12 % sulfuric acid by mass is:
А) 2.64; B) 1.32; C) 0.13; D) 0.26.
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5 . 4 . З а д а н и я д л я с а м о к о н т р о л я
1. Масса (г) медного купороса CuSO4 .5H2O, необходимого для приготовления 400 г 2 %-ного раствора CuSO4, равна:
А) 8; |
Б) 4.5; |
В) 12.5; |
Г) 80. |
2. Масса (г) карбоната калия, необходимого для приготовления одного
литра 0.6 М раствора |
K2CO3, составляет: |
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А) 230; |
Б) 83; |
В) 8.3; |
Г) 64. |
3. К 25 мл 85.16 %-ного раствора H2SO4 (ρ = 1.78 г/мл) прибавили 100 мл воды. Процентная концентрация серной кислоты в полученном растворе:
А) 2.6; Б) 38; В) 8.3; Г) 26.2.
4. Молярность (М) 70 % -ного раствора серной кислоты (ρ = 1.615 г/мл)
составляет: |
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А) 0.011; |
Б) 4.44; |
В) 11.5; |
Г) 1.15. |
5. Молярная концентрация (моль/л) 20 %-ного раствора NaOH, плотность которого 1.22 г/мл, составляет:
А) 0.61; Б) 0.006; В) 6.6; Г) 6.1.
6. Молярная концентрация (моль/л) 30 %-ного раствора азотной кислоты, плотность которого 1.18 г/мл, составляет:
А) 3.61; |
Б) 5.62; |
В) 22.3; |
Г) 0.56. |
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Молярная масса эквивалента сульфата хрома (III) составляет: |
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А) 392; |
Б) 65.3; |
В) 130.6; |
Г) 196. |
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Молярная масса эквивалента хлорида железа (III) составляет: |
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А) 27; |
Б) 162.5; |
В) 54.1; |
Г) 127. |
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Молярная масса эквивалента гидроксида алюминия составляет: |
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А) 234; |
Б) 39; |
В) 78; |
Г) 26. |
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10. Молярная концентрация химического эквивалента или нормальная концентрация 12 %-ного раствора серной кислоты (ρ = 1,08 г/мл) равна:
А) 2.64; Б) 1.32; В) 0.13; Г) 0.26.
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6 . F U N D A M E N T A L S O F C H E M I C A L
T H E R M O D Y N A M I C S
6 . 1 . D i c t i o n a r y |
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Boiling point |
Точка кипения |
Сalorimetry |
Калориметрия |
Closed system |
Закрытая система |
Complete combustion |
Полное сгорание |
Decomposition |
Разложение |
Endothermic reaction |
Эндотермическая реакция |
Enthalpy (H) |
Энтальпия (H) |
Entropy (S) |
Энтропия (S) |
Exothermic reaction |
Экзотермическая реакция |
From left to right |
Слева направо |
From right to left |
Справа налево |
Fusion |
Cплавление |
Vaporization |
Испарение |
Gibbs energy (G) |
Энергия Гиббса (G) |
Hess’s Law |
Закон Гесса |
Internal energy |
Внутренняя энергия |
Isolated system |
Изолированная система |
Melting point |
Точка плавления |
Open system |
Открытая система |
Specific heat capacity |
Удельная теплоемкость |
Standard enthalpy change of reac- |
Стандартное изменение энталь- |
tion ∆ Ho(298 K) |
пии реакции ∆Ho |
r |
298 реакц |
Standard enthalpy of formation |
Стандартная энтальпия образова- |
∆ Ho(298 K) |
ния ∆Ho |
f |
обр 298 |
Standard entropy So(298 K) |
Стандартная энтропия S298o |
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Standard entropy change of reaction |
Стандартное изменение энропии |
∆ So(298 K) |
реакции ∆So |
r |
298 реакц |
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Standard Gibbs energy change reac- |
Стандартное изменение энергии |
tion ∆ Go(298 K) |
Гиббса реакции ∆Go |
r |
298 реакц |
Standard Gibbs energy of formation |
Стандартная энергия Гиббса об- |
∆ Go(298 K) |
разования ∆Go |
f |
обр 298 |
Standard state |
Стандартное состояние |
State function |
Функция состояния |
Sublimation |
Возгонка |
Substance |
Вещество |
Thermodynamics |
Термодинамика |
Thermodynamically stable |
Термодинамически стабильный |
(unstable) |
(нестабильный) |
6 . 2 . W o r k e d e x a m p l e s
Example 1. Using data in table 6.1 calculate the standard enthalpy change (kJ) for the reaction
4NH3(g) + 3O2(g) = 2N2(g) + 6H2O(l).
Will this reaction be exothermic or endothermic?
Answer
The standard enthalpy change for a given reaction ∆fHo(298 K) is equal to the sum of the standard enthalpies of formation ∆fHo(298 K) of the products minus the sum of the standard enthalpies of formation of the reactants taking into account the stoichiometric coefficients.
∆rHo(298 K) = ∑ ∆fHo(products) − ∑ ∆fHo(reactants).
For the reaction
4NH3(g) + 3O2(g) = 2N2(g) + 6H2O(l)
∆rHo(298 K) = [ 2∆fHo(N2(g)) + 6∆fHo(H2O(l))] −
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− [4∆fHo(NH3(g)) + 3∆fHo(O2(g))].
The standard enthalpy of formation ∆fHo of a compound is defined as the change in enthalpy that accompanies the formation of 1 mole of a compound from its elements with all substances in their standard states. The value of the standard enthalpy of formation ∆fHo of some substances are given in Table 3. Since O2(g) and N2(g) are elements in its standard state, then their standard enthalpies of formation are zero.
From table 6.1
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∆fHo(H2O(l)) = –285.83 kJ/mol, ∆fHo(NH3(g)) = –46.2 kJ/mol.
Then
∆rHo(298 K) = 6 (–285.83) − 4 (–46.2) = –1530.18 kJ.
∆rHo < 0, therefore, this reaction is exothermic.
Example 2. Explain how the standard entropy, So(298 K) changes during the transition I2(s) → I2(g).
Answer
There is a general tendency for values of the standard entropy So(298 K) for solid substances to be smaller than those for liquids, while gases possess the largest standard entropy. Therefore So(298 K) (I2(s)) < So(298 K) (I2(g)). Indeed, according to the Table 2: So(298 K)(I2(s)) = 116.15 J/(K mol); So(298 K) (I2(g)) = 260.59 J/(K mol).
Example 3. Calculate the Gibbs energy change (kJ) at 80 oC for the reaction
4NO2(g) + О2(g) + 2Н2О(l) = 4HNO3(l),
if ∆rHo(298 K) = –261.5 kJ; ∆rSo(298 K) = –199.2 J/K.
What factor (enthalpy or entropy) favors the process of the HNO3 formation?
Answer
The change in Gibbs energy of a reaction at a given temperature is given by the equation:
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