Questions for self-control. Introduction to general chemistry Examples of problem solving

What are solutions and what characteristics 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 called? Define percentage, molar, molar equivalent and molal concentrations, and 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 10 g of potassium nitrate (KNO 3) was dissolved in 240 g of water, the temperature of the solution decreased by 3.4 degrees. Determine the heat of solution of the salt. The specific heat capacity (sp) 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) = - 3556.4 J = - 3.56 kJ.

3. We calculate the amount of heat absorbed when dissolving one mole of KNO 3, i.e. its heat of dissolution (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 solution of KNO 3 is 35.96 kJ/mol.

Solution:

1. 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 solution:

m =  . V = 1.12 . 1000 = 1120 g;

b) find the weight amount of sulfuric acid:

100 g of solution contains 17.5 g of H 2 SO 4;

in 1120 g of solution - x,

2. Find the titer of the solution; To do this, it is necessary to divide the weight amount of acid contained in a known volume of solution by the volume of solution expressed in milliliters:

T = = 0.196 g/ml.

3. Find the molar concentration of the solution; To do 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; To do this, it is necessary to divide the weight of the 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 mol equiv/l.

The molar concentration equivalent can also be calculated using the formula

.

5.Calculate the molality of the solution; To do this, you need to find the number of moles of acid contained in 1000 g of solvent (water).

From previous calculations (see point 3) it is known that 1120 g (1 l) of solution contains 196 g or 2 moles of H2SO4, therefore, water in such a solution:

1120 - 196 = 924 g.

Let's make a proportion:

per 924 g of water there are 2 moles of H 2 SO 4

per 1000 g of water - x.

With m = x = = 2.16 mol/1000 g of water.

Answer: T = 0.196 g/ml; = 2 mol/l; C eq = 4 mol equiv/l;

With m = 2.16 mol/1000 g of water.

Example 3. How many milliliters of a 96% solution of H 2 SO 4 ( = 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 solution with a molar equivalent concentration of 0.5 (the equivalent of sulfuric acid is 49 g):

1000 ml of 0.5 N solution contains 49. 0.5 = 24.5 g H 2 SO 4.

2. Determine the weight amount of the original (96%) 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 original solution by dividing the weight amount of the solution by its density ():

V = = 13.87 ml.

Answer: to prepare 1 liter of sulfuric acid solution with a molar concentration equivalent of 0.5, 13.87 ml of a 96% solution of H 2 SO 4 is required.

Example 4. A solution prepared from 2 kg (m) of ethyl alcohol and 8 kg (g) of water was poured into the car radiator. Calculate the freezing point of the solution. The cryoscopic water constant Kk is 1.86.

Solution.

1. Find the decrease in the freezing temperature of the solution using a corollary from Raoult’s law:

т з = K к С m = K к .

The molar mass of C 2 H 5 OH is 46 g/mol, therefore,

Т з = 1.86 = 10.1 о С.

2. Find the freezing temperature of the solution:

T s = 0 - 10.1 = - 10.1 o C.

Answer: the solution freezes at a temperature of -10.1 o C.

Properties of dilute solutions that depend only on the amount of nonvolatile solute are called colligative properties. These include a decrease in the vapor pressure of the solvent above the solution, an increase in the boiling point and a decrease in the freezing point of the solution, as well as osmotic pressure.

Decreasing the freezing point and increasing the boiling point of a solution compared to a pure solvent:

T deputy = = K TO. m 2 ,

T kip. = = K E. m 2 .

Where m 2 – molality of the solution, K K and K E – cryoscopic and ebullioscopic solvent constants, X 2 – mole fraction of solute, H pl. And H Spanish – enthalpy of melting and evaporation of the solvent, T pl. And T kip. – melting and boiling points of the solvent, M 1 – molar mass of the solvent.

Osmotic pressure in dilute solutions can be calculated using the equation

Where X 2 is the molar fraction of the solute, and is the molar volume of the solvent. In very dilute solutions this equation becomes van't Hoff equation:

Where C– molarity of the solution.

Equations that describe the colligative properties of nonelectrolytes can also be applied to describe the properties of electrolyte solutions by introducing the Van't Hoff correction factor i, For example:

= iCRT or T deputy = iK TO. m 2 .

The isotonic coefficient is related to the degree of electrolyte dissociation:

i = 1 + ( – 1),

where is the number of ions formed during the dissociation of one molecule.

Solubility of a solid in an ideal solution at temperature T described Schroeder equation:

,

Where X– mole fraction of solute in solution, T pl. – melting temperature and H pl. – enthalpy of melting of the solute.

EXAMPLES

Example 8-1. Calculate the solubility of bismuth in cadmium at 150 and 200 o C. The enthalpy of fusion of bismuth at the melting temperature (273 o C) is 10.5 kJ. mol –1 . Assume that an ideal solution is formed and the enthalpy of fusion does not depend on temperature.

Solution. Let's use the formula .

At 150 o C , where X = 0.510

At 200 o C , where X = 0.700

Solubility increases with temperature, which is characteristic of an endothermic process.

Example 8-2. A solution of 20 g of hemoglobin in 1 liter of water has an osmotic pressure of 7.52 10 –3 atm at 25 o C. Determine the molar mass of hemoglobin.

65 kg. mol –1 .

TASKS

1. Calculate the minimum osmotic work performed by the kidneys to excrete urea at 36.6 o C, if the concentration of urea in plasma is 0.005 mol. l –1, and in urine 0.333 mol. l –1.
2. 10 g of polystyrene is dissolved in 1 liter of benzene. The height of the solution column (density 0.88 g cm–3) in the osmometer at 25 o C is 11.6 cm. Calculate the molar mass of polystyrene.
3. Human serum albumin protein has a molar mass of 69 kg. mol –1 . Calculate the osmotic pressure of a solution of 2 g of protein in 100 cm 3 of water at 25 o C in Pa and in mm of the solution column. Assume the density of the solution to be 1.0 g cm–3.
4. At 30 o C, the vapor pressure of an aqueous solution of sucrose is 31.207 mm Hg. Art. The vapor pressure of pure water at 30 o C is 31.824 mm Hg. Art. The density of the solution is 0.99564 g cm–3. What is the osmotic pressure of this solution?
5. Human blood plasma freezes at -0.56 o C. What is its osmotic pressure at 37 o C, measured using a membrane permeable only to water?
6. *The molar mass of the enzyme was determined by dissolving it in water and measuring the height of the solution column in an osmometer at 20 o C, and then extrapolating the data to zero concentration. The following data was received:

C, mg. cm –3
h, cm

7. The molar mass of a lipid is determined by the increase in boiling point. The lipid can be dissolved in methanol or chloroform. The boiling point of methanol is 64.7 o C, the heat of evaporation is 262.8 cal. g –1 . The boiling point of chloroform is 61.5 o C, the heat of evaporation is 59.0 cal. g –1 . Calculate the ebullioscopic constants of methanol and chloroform. Which solvent is best to use to determine molar mass with maximum accuracy?
8. Calculate the freezing point of an aqueous solution containing 50.0 g of ethylene glycol in 500 g of water.
9. A solution containing 0.217 g of sulfur and 19.18 g of CS 2 boils at 319.304 K. The boiling point of pure CS 2 is 319.2 K. The ebullioscopic constant of CS 2 is 2.37 K. kg. mol –1 . How many sulfur atoms are there in a sulfur molecule dissolved in CS 2?
10. 68.4 g of sucrose dissolved in 1000 g of water. Calculate: a) vapor pressure, b) osmotic pressure, c) freezing point, d) boiling point of the solution. The vapor pressure of pure water at 20 o C is 2314.9 Pa. Cryoscopic and ebullioscopic constant waters are 1.86 and 0.52 K. kg. mol –1 respectively.
11. A solution containing 0.81 g of hydrocarbon H(CH 2) nH and 190 g of ethyl bromide freezes at 9.47 o C. The freezing point of ethyl bromide is 10.00 o C, the cryoscopic constant is 12.5 K. kg. mol –1 . Calculate n.
12. When 1.4511 g of dichloroacetic acid is dissolved in 56.87 g of carbon tetrachloride, the boiling point increases by 0.518 degrees. Boiling point CCl 4 76.75 o C, heat of evaporation 46.5 cal. g –1 . What is the apparent molar mass of the acid? What explains the discrepancy with the true molar mass?
13. A certain amount of a substance dissolved in 100 g of benzene lowers its freezing point by 1.28 o C. The same amount of a substance dissolved in 100 g of water lowers its freezing point by 1.395 o C. The substance has a normal molar mass in benzene, and in water completely dissociated. How many ions does a substance dissociate into in an aqueous solution? The cryoscopic constants for benzene and water are 5.12 and 1.86 K. kg. mol –1 .
14. Calculate the ideal solubility of anthracene in benzene at 25 o C in molality units. The melting enthalpy of anthracene at the melting point (217 o C) is 28.8 kJ. mol –1 .
15. Calculate solubility P-dibromobenzene in benzene at 20 and 40 o C, assuming that an ideal solution is formed. Melting enthalpy P-dibromobenzene at its melting point (86.9 o C) is 13.22 kJ. mol –1 .
16. Calculate the solubility of naphthalene in benzene at 25 o C, assuming that an ideal solution is formed. The melting enthalpy of naphthalene at its melting temperature (80.0 o C) is 19.29 kJ. mol –1 .
17. Calculate the solubility of anthracene in toluene at 25 o C, assuming that an ideal solution is formed. The melting enthalpy of anthracene at the melting point (217 o C) is 28.8 kJ. mol –1 .
18. Calculate the temperature at which pure cadmium is in equilibrium with a Cd – Bi solution, the mole fraction of Cd in which is 0.846. The melting enthalpy of cadmium at the melting point (321.1 o C) is 6.23 kJ. mol –1 .

Problem 427.
Calculate the mole fractions of alcohol and water in a 96% (by weight) solution of ethyl alcohol.
Solution:
Mole fraction(N i) – the ratio of the amount of 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 alcohol; n 2 - amount of water.

Let's calculate the mass of alcohol and water contained in 1 liter of solution, provided that their densities are equal to one of the proportions:

a) mass of alcohol:

b) mass of water:

We find the amount of substances using the formula: , where m(B) and M(B) are the mass and amount of the substance.

Now let's calculate the mole fractions of substances:

Answer: 0,904; 0,096.

Problem 428.
666 g of KOH 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:
A) 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) = m(H 2 O) + m(KOH) = 1000 + 666 = 1666 g.

b) Molar (volume-molar) concentration shows the number of moles of solute contained in 1 liter of solution.

Let's find the mass of KOH per 100 ml of solution using 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 g.

Now let's calculate the molarity of the solution:

We find how many grams of HNO 3 are per 1000 g of water by making the proportion:

d) Mole fraction (Ni) – the ratio of the amount of dissolved substance (or solvent) to the sum of the amounts of all substances in the solution. In a system consisting of alcohol and water, the mole fraction of water (N 1) is equal to the mole fraction of alcohol, where n 1 is the amount of alkali; n 2 - amount of water.

100g of this solution contains 40g KOH and 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 mass) H 2 SO 4 solution is 1.105 g/ml. Calculate: a) normality; b) molarity; c) molality of the solution.
Solution:
Let's find the mass of the solution using the formula: m = p V, where p- density of the solution, V - volume of the solution.

m(H 2 SO 4) = 1.105 . 1000 = 1105 g.

The mass of H 2 SO 4 contained in 1000 ml of solution is found from the proportion:

Let us determine the molar mass of the equivalent of H 2 SO 4 from the relationship:

ME (V) - molar mass of 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) shows the number of equivalents of a solute contained in 1 liter of solution.

b) Molal concentration

Now let's calculate the molality of the solution:

c) Molal concentration (or molality) shows the number of moles of 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 let's 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) sucrose solution C 12 H 22 O 11 is 1.035 g/ml. Calculate: a) the concentration of sucrose in g/l; b) molarity; c) molality of the solution.
Solution:
M(C 12 H 22 O 11) = 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) = 1.035. 1000 = 1035 g.

a) We calculate the mass of C 12 H 22 O 11 contained in the solution using the formula:

Where
- mass fraction of dissolved substance; m (in-va) - mass of dissolved substance; m (solution) - mass of solution.

The concentration of a substance in g/l shows the number of grams (units of mass) contained in 1 liter of solution. Therefore, the concentration of sucrose is 93.15 g/l.

b) Molar (volume-molar) concentration (CM) shows the number of moles of a dissolved substance contained in 1 liter of solution.

V) Molal concentration(or molality) shows the number of moles of 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 let's calculate the molality of the solution:

Answer: a) 93.15 g/l; b) 0.27 mol/l; c) 0.29 mol/kg.