Tasks for solutions and methods for their solution

Solving problems for solutions is an important part of chemistry in a modern school. Many children have some difficulties in performing calculations, due to the lack of understanding of the sequence of the task. Let us analyze some terms that include tasks for solutions in chemistry, and give examples of ready-made solutions.

Percent Concentration

Tasks assume the composition and solution of the proportion. Taking into account that this type of concentration is expressed in mass fractions, it is possible to determine the content of the substance in the solution.

This quantity is a quantitative characteristic of the solution proposed in the problem. Depending on the type of task, it is necessary to determine the new percentage concentration, calculate the mass of the substance, calculate the volume of the solution.

The molar concentration

Some problems on the concentration of solutions are associated with the determination of the amount of substance in the volume of the solvent. The unit of measurement of this value is mol / l.

In the school curriculum, assignments of this kind are found only at the higher stage of education.

Singularities of problems on solutions

Let's present some problems on solutions in chemistry with a solution to show the sequence of actions during their analysis. First, we note that you can do drawings to understand the essence of the processes described in the proposed task. If desired, you can make out the task in the form of a table, in which the initial and sought-for values will be supplied.

Task 1

A container containing 5 liters of a 15% solution of salt was poured into seven liters of water. Determine the percentage concentration of the substance in the new solution.

In order to determine the required quantity, we denote it by X. We calculate the quantitative content of the substance in the first solution through the proportion: if 5 is multiplied by 0.15, we obtain 0.75 grams.

Next, we calculate the mass of the new solution, taking into account that 7 liters of water were poured in, and we get 12 grams.

We find the content in percentage of table salt in the resulting solution based on the definition of this value, we obtain: (0.75: 12) x 100% = 6.25%

Here is another example of a task related to the use of mathematical proportions in calculations.

Task 2

How much by weight of copper should be added to a piece of bronze, having a mass of 8 kilograms, containing 13 percent of pure metal, to increase the percentage of copper to 25 percent.

Such solution problems first require the determination of the mass of pure copper in the initial alloy. For this you can use the mathematical proportion. As a result, it turns out that the mass is: 8 x 0.13 = 1.04 kg

We take the desired value for x (grams), then in the new alloy we get its value (1.04 + x) kilograms. We express the mass of the alloy obtained, we obtain: (8 + x) kilograms.

In the problem, the percentage of metal in the new alloy is 25 percent, and a mathematical equation can be made.

A variety of solution tasks are included in test tasks to test the level of subject knowledge of graduates of the eleventh grades. We give some conditions and solutions of problems of this type.

Task 3

Determine the volume (under normal conditions) of the gas that was collected after the introduction of 0.3 moles of pure aluminum in 160 milliliters of a warm 20% solution of potassium hydroxide (1.19 g / ml).

The sequence of calculations in this task:

1. First you need to determine the mass of the solution.
2. Next, the amount of alkali is calculated.
3. The obtained parameters are compared among themselves, the shortage is determined. Subsequent calculations are carried out on a substance taken in insufficient quantities.
4. We write the equation of the reaction that occurs between the initial substances, we arrange the stereochemical coefficients. We perform the calculations by the equation.

The weight of the alkali solution used in the task is 160 x 1.19 = 190.4 g.

The mass of the substance will be 38.08 grams. The amount of alkali taken is 0.68 mol. In the condition it is said that the amount of aluminum is 0.3 mole, therefore, this metal is present in a deficiency.

The subsequent calculations are carried out on it. It turns out that the volume of gas will be 0.3 x 67.2 / 2 = 10.08 liters.

Problems for solutions of this type in graduates cause maximum difficulties. The reason for the inadequacy of the sequence of actions, as well as in the absence of formed representations about basic mathematical calculations.

Task 4

Problems on the topic "Solutions" can include the determination of a pure substance at a given percentage of impurities. Let's give an example of such a task, so that the guys did not have any difficulties with its implementation.

Calculate the volume of gas obtained by the action of concentrated sulfuric acid on 292.5 g of salt with 20% impurities.

Sequencing:

1. Taking into account that 20% of impurities are present in the condition of the problem, it is necessary to determine the content of the substance by mass (80%).
2. We prescribe the equation of the chemical reaction, we arrange the stereochemical coefficients. We calculate the volume of the evolved gas using the molar volume.

The mass of the substance, based on the fact that there are impurities, yields 234 grams. And when carrying out calculations on this equation, we get that the volume will be 89.6 liters.

Problem 5

What else is proposed in the school curriculum for chemistry of the problem of solutions? Here is an example of a task related to the need to calculate the mass of a product.

Lead sulphide (II) having a mass of 95.6 grams reacts with 300 milliliters of a 30% hydrogen peroxide solution (density: 1.1222 g / ml). The reaction product is (in grams) ...

The order of solving the problem:

1. Solutions of substances we translate through proportions into mass.
2. Next, determine the amount of each source component.
3. After comparing the results, select the substance that is taken in insufficient quantities.
4. Calculations are carried out for the substance taken in the defect.
5. We compose the equation of chemical interaction and calculate the mass of the unknown substance.

We calculate the solution of peroxide, it is 336.66 grams. The mass of the substance will correspond to 100,99 grams. Calculate the number of moles, it will be 2.97. Lead sulphide will be 95.6 / 239 = 0.4 mol, (it is contained in the deficit).

We form the equation of chemical interaction. We determine the desired value from the scheme and obtain 121.2 grams.

Task 6

Find the amount of gas (mol) that can be obtained by thermal firing of 5.61 kg of iron (II) sulphide having a purity of 80%.

Procedure:

1. We calculate the mass of pure FeS.
2. We write down the equation of its chemical interaction with air oxygen. We perform calculations on the reaction.

By mass, the pure substance will amount to 4488 g. The amount of the component to be determined will be 51 liters.

Problem 7

A solution was prepared from 134.4 liters (under normal conditions) of sulfur oxide (4). To it were added 1.5 liters of a 25% solution of sodium hydroxide (1.28 g / ml). Determine the mass of the resulting salt.

The algorithm of calculations:

1. We calculate the mass of the alkali solution by the formula.
2. We find the mass and the number of moles of caustic soda.
3. We calculate the same value for sulfur oxide (4).
4. By the ratio of the obtained indices, we determine the composition of the salt formed, we determine the defect. Calculations are carried out for lack.
5. We record the chemical reaction with the coefficients, calculate the mass of the new salt by the defect.

As a result, we get:

• Solution of alkali will be 1171.875 grams;
• By weight of sodium hydroxide will be 292.97 grams;
• In moles of this substance contains 7.32 mol;
• Analogously, we calculate for sulfur oxide (4), we obtain 6 moles;
• As a result of the interaction, an average salt will be formed;
• We get 756 grams.

Task 8

To 100 grams of a 10% solution of ammonium chloride, 100 g of a 10% solution of silver nitrate were added. Determine the mass (in grams) of the precipitate.

The algorithm of calculations:

1. We calculate the mass and the amount of ammonium chloride material.
2. We calculate the mass and amount of the substance of the salt - silver nitrate.
3. We determine which of the initial substances was taken in insufficient quantities, we carry out calculations on it.
4. We write down the equation of the occurring reaction, we carry out calculations of the mass of the precipitate.

Ammonium chloride by weight will be 10 g, in quantity - 0.19 mole. Silver nitrate is taken 10 grams, which is 0.059 mole. When calculating the shortfall, we get a mass of salt of 8.46 grams.

In order to cope with the complex tasks that are offered at the final exams in the ninth and eleventh grade (at the rate of unlimited chemistry), you need to own algorithms and have certain computing skills. In addition, it is important to know the technology of composing chemical equations, to be able to arrange the coefficients in the process.

Without such elementary skills and skills, even the simplest task of determining the percentage concentration of a substance in a solution or mixture will seem to the graduate a difficult and serious test.