True solutions ways to express the concentration of solutions presentation. Methods of expressing the concentration of solutions

§ 1 Methods of expressing the concentration of solutions

Let's try to imagine two solutions. One is prepared by dissolving one tablespoon of salt in a glass of water, the second by dissolving a tablespoon of salt in a bath of water.

Will these solutions be different? Of course yes. The first solution will taste saltier, boil at a higher temperature, and freeze at a lower temperature than the second solution. And chemical reactions with the first solution will proceed more intensely than with the second. Thus, the ratio of the amount of solute to the amount of solvent (that is, concentration) determines the properties of the solution.

Solutions are concentrated (with high content solute) and dilute (low solute content).

This qualitative assessment concentration of solutions, which can be used very conditionally. Of much greater interest are the various quantitative methods expressions for the concentration of solutions.

The concentration of a substance can be expressed in moles of dissolved substance per 1 liter of solution. This concentration is called molar and is denoted in Latin capital letter WITH.

Molar concentration C is equal to the ratio of the amount of substance in moles υ to the volume of solution in liters V. It is expressed in moles per liter.

The concentration of a solution is often expressed in mass fractions.

The mass fraction of a solute is the ratio of the mass of the solute to the total mass of the solution.

The mass fraction of the dissolved substance is denoted by Greek letter ω.

The mass fraction ω of a substance is equal to the ratio of the mass of the substance m to the mass of the solution mр. The mass fraction can be expressed as a fraction of a unit or as a percentage, in which case the result is multiplied by 100%.

§ 2 Solving problems on the topic of the lesson

Let's solve the problem. After complete evaporation of 50 g of solution, 6 g of a solid residue was formed. Calculate the mass fraction of the solute in the solution taken.

As specified mass fraction calculate the quantities of ingredients for preparing solutions.

For example, you need to prepare 150 g of a 10% sodium chloride solution, that is, you need to find how much salt and water is needed for this purpose.

Answer: to prepare the solution you will need 15 g of sodium chloride and 135 g of water.

List of used literature:

NOT. Kuznetsova. Chemistry. 8th grade. Tutorial for educational institutions. – M. Ventana-Graf, 2012.

Images used:

Mass fraction The ratio of the mass of the solute to the total mass of the solution. The mass fraction is expressed in fractions of a unit: w(solv. in - va) = m (solv. in - va)/ m (p - ra) or as a percentage: w(solv. in - va) = m (solv. in - va)/ m (p - ra) * 100% Problem

Solve the problems Determine the mass fraction (in%) of NaOH in the solution if NaOH weighing 16 g is dissolved in water weighing 144 g. How much salt and water will be required to prepare 200 g of 15% sodium carbonate solution. When evaporating 25 g of solution, 0.25 g of salt was obtained. Determine the mass fraction of the dissolved substance and express it as a percentage. 500 g of water was added to 200 g of a 20% solution. What is the mass fraction of solute in the resulting solution?

Unified State Exam. Task B 9. The mass of potassium nitrate that should be dissolved in 150 g of a solution with a mass fraction of 10% to obtain a solution with a mass fraction of 12% is equal to: __________ g. (Write down the number to the nearest tenth.) Determine the mass of water that must be added to 20 g of solution acetic acid with a mass fraction of 70% to obtain a solution of vinegar with a mass fraction of 3%. Answer: __________ (Write the number to the nearest whole number.)

Reception of the “cross” In what mass ratio should the 5% and 60% solutions be mixed to prepare 200 g of a 20% solution. GIVEN: W1=5% W2=60% W = 20% m = 200 g m1/m2 -? SOLUTION: We draw up a diagonal diagram: in the center we write the required mass fraction. At the left end of each diagonal we write these mass fractions. Then we subtract diagonally (we always subtract from larger size smaller): = = 15 We put the result of the subtraction at the right end of the corresponding diagonal: Thus, 60% and 5% solutions should be mixed in the ratio 15:40 = 3:8. Total 3+8 = 11 parts by mass. total weight solution should be equal to 200 g. Therefore, 1 m h is 200 g / 11 = 18.18 g. Therefore, 3 m h will be 18.18 g x 3 = 54.54 g, and 8 m h. - 18.18 g x 8 = 145.46 g. Answer: you need to take 54.54 g of 60% solution and 145.5 g of 5%.

Solve the problem (reception of the “cross”) In what mass ratio should the 3% and 40% solutions be mixed to obtain 150 g of a 15% solution? From the Unified State Examination in mathematics There are two alloys. The first alloy contains 10% nickel, the second 30% nickel. From these two alloys, a third alloy weighing 200 kg was obtained, containing 25% nickel. How many kilograms is the mass of the first alloy less than the mass of the second?

Unified State Exam. Mathematics 7 liters of water were added to a vessel containing 5 liters of a 12% aqueous solution of a certain substance. What percentage is the concentration of the resulting solution? We mixed a certain amount of a 15 percent solution of a certain substance with the same amount of a 19 percent solution of this substance. What percentage is the concentration of the resulting solution?

Unified State Exam. Mathematics Mixing 30 percent and 60 percent acid solutions and adding 10 kg clean water, obtained a 36 percent acid solution. If instead of 10 kg of water we added 10 kg of a 50 percent solution of the same acid, we would get a 41 percent acid solution. How many kilograms of a 30 percent solution were used to obtain the mixture? Grapes contain 90 moisture content, and raisins contain 5. How many kilograms of grapes are required to produce 20 kilograms of raisins?

Molar concentration Molar concentration c (solv. in - va) is the ratio of the amount of substance n (mol) contained in a solution to the volume of this solution V (l): c (solv. in - va) = m(solv. in - va )/M(solv. in - va)*V, since n = m / M, then c (solv. in - va) = n (solv. in - va)/V

Molar Concentration The symbol “M” is used to indicate molar concentration. If 1 liter of solution contains 1 mol of solute, then the solution is called unipolar and is designated 1 M, if 2 mol is bipolar (designated 2 M), 0.1 mol is decimolar (0.1 M) Problems

Solve the problems How many grams of H 2 SO 4 are contained in a 0.1 M solution with a volume of 500 ml? Calculate the molar concentration of a sodium hydroxide solution, 1 liter of which contains 20 g of NaOH? Problems 17, 18, 19 page 64 (Collection of problems and exercises in chemistry, author Yu. M. Erokhin)

Download:

Preview:

To use presentation previews, create an account for yourself ( account) Google and log in: https://accounts.google.com

Slide captions:

What do these quantities mean in chemistry? ω, cm, X

Lesson topic: “Ways to express the concentrations of solutions”

Lesson objectives: expand and systematize ideas about ways to express the concentration of solutions; explore new ways of expressing the concentrations of solutions; learn to apply what you have learned theoretical knowledge when solving problems; develop intellectual skills and abilities.

Concentration is a value characterizing quantitative composition solution. According to the IUPAC rules, the concentration of a dissolved substance (not a solution) is the ratio of the amount of a dissolved substance or its mass to the volume of a solution (mol/l, g/l), that is, this is the ratio of heterogeneous quantities. Those quantities that are the ratio of similar quantities (the ratio of the mass of a dissolved substance to the mass of a solution, the ratio of the volume of a dissolved substance to the volume of a solution) are correctly called fractions. However, in practice, for both types of expression of composition, the term concentration is used and they speak of the concentration of solutions.

Methods of expressing the concentration of solutions 1 Mass fraction (weight percentage, percentage concentration) 2 Volume fraction 3 Molarity (molar concentration) 4 Mole fraction 5 Molality (molal concentration) 6 Solution titer 7 Normality (molar concentration equivalent) 8 Solubility of a substance

Percent concentration, mass fraction of solute Mass fraction of solute is the ratio of the mass of the solute to the mass of the solution. To calculate the percentage concentration, the formula is used: A solution consists of a solute and a solvent. The mass of the solution can be determined by the formula:

In binary solutions there is often a clear relationship between the density of the solution and its concentration (at a given temperature). This makes it possible to determine concentrations in practice important solutions using a densimeter (alcohol meter, saccharimeter, lactometer). Some hydrometers are calibrated not in density values, but directly in the concentration of the solution (alcohol, fat in milk, sugar). Often, to express concentration (for example, sulfuric acid in batteries), they simply use their density. Hydrometers designed to determine the concentration of solutions of substances are common.

Dependence of the density of H 2 SO 4 solutions on its mass fraction in aqueous solution at 20°C ω, % 10 30 50 70 80 90 ρ H 2 SO 4, g/ml 1.066 1.219 1.395 1.611 1.727 1.814

Volume fraction Volume fraction is the ratio of the volume of a dissolved substance to the volume of a solution. The volume fraction is measured in fractions of a unit or as a percentage. where: V (v-va) - volume of dissolved substance, l; V(r-ra) - overall volume solution, l. As mentioned above, there are hydrometers designed to determine the concentration of solutions of certain substances. Such hydrometers are calibrated not in density values, but directly in the concentration of the solution. φ = V(in-va) V(r-ra)

Molarity (molar concentration) Molarity is the number of moles of solute per unit volume of solution. where ν is the amount of dissolved substance, mol; V - volume of solution, l Molarity is often expressed in mol/l or mmol/l. The following designations for molar concentration are possible: C, Cm, M. Thus, a solution with a concentration of 0.5 mol/l is called 0.5 molar (0.5 M).

Mole fraction Mole fraction (X) is the ratio of the number of moles of a given component to the total number of moles of all components. The mole fraction is expressed in fractions of a unit. X = ν (amount) \ ∑ ν (amount) ν - amount of component, mol; ∑ ν - the sum of the quantities of all components, mol.

Molality (molal concentration) Molality is the number of moles of dissolved substance in 1 kg of solvent. It is measured in mol/kg. Thus, a solution with a concentration of 0.5 mol/kg is called 0.5-molal. St = ν / m(p-la), where: ν - amount of dissolved substance, mol; m (r-la) - mass of solvent, kg. Should be paid Special attention, that despite the similarity of names, molarity and molality are different quantities. First of all, when expressing concentration in molality, as opposed to molarity, the calculation is based on the mass of the solvent, and not on the volume of the solution. Molality, unlike molarity, does not depend on temperature.

Solution titer Solution titer is the mass of the dissolved substance in 1 ml of solution. T= m (volume)/ V (solution), where: m (volume) - mass of dissolved substance, g; V(solution) - total volume of solution, ml; IN analytical chemistry Usually the titrant concentration is recalculated in relation to specific reaction titration in such a way that the volume of titrant used directly shows the mass of the substance being determined; that is, the titer of the solution shows what mass of the analyte (in grams) corresponds to 1 ml of the titrated solution.

Normality (molar concentration equivalent) Normality (Сн) - number of equivalents of this substance in one liter of solution. Normality is expressed in mol-eq/l. Often the concentration of such solutions is expressed as “n”. For example, a solution containing 0.1 mol-eq/l is called decinormal and is written as 0.1 N. CH = E/ V (solution), where: E - equivalent, mol-equiv; V - total volume of solution, l; CH(alkalis) ∙V(alkalis)= CH(acids)∙V(acids)

Solubility coefficient Very often, the concentration of a saturated solution, along with the above characteristics, is expressed through the so-called solubility coefficient or simply the solubility of the substance. The ratio of the mass of a substance forming a saturated solution at a given temperature to the mass of the solvent is called the solubility coefficient: Kp = m (v-va) / m (p-la) The solubility of a substance shows the maximum mass of a substance that can dissolve in 100 g of solvent: p = (m v-va / m r-la) ∙ 100%

Problems 1. Determine the molar concentration of sodium chloride in a 24% solution with a density of 1.18 g/ml. (Answer - 4.84 M) 2. Determine the molar concentration of hydrochloric acid in a 20% solution with a density of 1.098. (Answer - 6M) 3. Determine the molar concentration nitric acid in a 30% solution with a density of 1.18 g/ml. (Answer - 5.62 M) 4. Calculate the mass fraction of potassium hydroxide in an aqueous solution with a concentration of 3 M and a density of 1.138 g/ml. (Answer - 15%) 5. How many ml of 56% sulfuric acid solution (density 1.46 g/ml) is needed to prepare 3 liters of 1M solution? (Answer - 360 ml)

6. A 2M solution of potassium chloride with a volume of 40 ml and a density of 1.09 g/ml was added to water weighing 200 g. Determine the molar concentration and mass fraction of salt in the resulting solution if its density turns out to be 1.015 g/ml. (Answer - 0.33M, 2.45%) 7. How many g of potassium hydroxide is needed to neutralize 300 ml of 0.5 M sulfuric acid solution? (Answer - 16.8 g) 8. What volume of a 2 M solution of potassium hydroxide will react: a) with 49 g of sulfuric acid b) with 200 g of a 24.5% solution of sulfuric acid? C) with 50 g of 6.3% nitric acid solution? 9. What volume of a 3M sodium chloride solution with a density of 1.12 g/ml must be added to water weighing 200 g to obtain a solution with a mass fraction of salt of 10%? (Answer - 315 ml) 10. What volume of 3M potassium chloride solution will be required to prepare 200 ml of 8% salt solution with a density of 1.05 g/ml? (Answer - 75.2 ml) 11. 2.8 liters of ammonia were dissolved in water, the volume of the solution was brought to 500 ml. What amount of ammonia is contained in 1 liter of such a solution? (Answer - 0.25 mol)

True solutions ways to express the concentration of solutions presentation. Methods of expressing the concentration of solutions

Similar documents

Download:

Preview:

Slide captions: