Solving problems in solutions is an important section of a chemistry course in a modern school. Many children have certain difficulties in performing the calculations associated with the lack of ideas about the sequence of the task. Let us analyze some terms that include tasks on solutions in chemistry and give examples of ready-made solutions.
Percentage concentration
Tasks involve the preparation and solution of the proportion. Given that this type of concentration is expressed in mass fractions, it is possible to determine the content of a substance in a solution.
The mentioned value is a quantitative characteristic of the solution proposed in the problem. Depending on the type of task, it is necessary to determine a new percentage concentration, calculate the mass of the substance, calculate the volume of the solution.
Molar concentration
Some problems in the concentration of solutions are associated with determining the amount of a substance in the volume of a solvent. The units of this value are mol / L.
In the school curriculum, tasks of this kind are found only at the senior level of training.
Features of solutions to solutions
We present some problems for chemistry solutions with a solution to show the sequence of actions during their analysis. To begin with, we note that you can make drawings to understand the essence of the processes described in the proposed task. If you wish, you can draw up the task in the form of a table in which the initial and desired values will be set.
Task 1
Seven liters of water were poured into a container containing 5 liters of a 15% salt solution. Determine the percentage concentration of the substance in the new solution.
In order to determine the desired value, we denote it by X. Using the proportion, we calculate the quantitative content of the substance in the first solution: if 5 times 0.15, we get 0.75 grams.
Next, we calculate the mass of the new solution, given that 7 liters of water were poured, and we get 12 grams.
We find the percentage in table salt in the resulting solution based on the definition of this value, we obtain: (0.75: 12) x 100% = 6.25%
We give one more example of a task connected with the use of mathematical proportions in calculations.
Task 2
How much by weight of copper must be added to a piece of bronze having a mass of 8 kilograms containing 13 percent pure metal in order to increase the percentage of copper to 25%.
Such solutions to solutions first require determining the mass of pure copper in the initial alloy. To do this, you can use the mathematical proportion. The result is 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. Express the mass of the resulting alloy, we get: (8 + x) kilograms.
In the problem, the percentage of metal in the new alloy is 25 percent; a mathematical equation can be made.
A variety of tasks for solutions are included in test tasks to check the level of subject knowledge of graduates of the eleventh grades. We give some conditions and solutions to tasks of this type.
Task 3
Determine the volume (under normal conditions) of the gas that was collected after the introduction of 0.3 mol of pure aluminum in 160 milliliters of a warm 20% potassium hydroxide solution (1.19 g / ml).
The sequence of calculations in this task:
- First you need to determine the mass of the solution.
- Next, the amount of alkali is calculated.
- The obtained parameters are compared with each other, shortage is determined. Subsequent calculations are carried out on a substance taken in insufficient quantities.
- We write the equation of the reaction occurring between the starting materials, arrange the stereochemical coefficients. We carry out calculations by the equation.
The mass of the alkali solution used in the problem 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. The condition says that the amount of aluminum is 0.3 mol, therefore, this metal is present in the deficiency.
Subsequent calculations are carried out precisely on it. It turns out that the gas volume will be 0.3 x 67.2 / 2 = 10.08 liters.
Problems with solutions of this type among graduates cause maximum difficulties. The reason is the unworked sequence of actions, as well as the lack of formed ideas about basic mathematical calculations.
Task 4
Tasks on the topic “Solutions” may include the determination of a pure substance at a given percentage of impurities. We give an example of such a task so that the guys do not have any difficulties with its implementation.
Calculate the volume of gas produced by exposure to concentrated sulfuric acid on 292.5 g of salt with 20% impurities.
Sequencing:
- Given that the condition of the problem states the presence of 20 percent of impurities, it is necessary to determine the substance content by weight (80%).
- We write the chemical reaction equation, arrange the stereochemical coefficients. We calculate the volume of gas released using the molar volume.
The mass of the substance, based on the fact that there are impurities, is 234 grams. And when carrying out calculations according to this equation, we get that the volume will be equal to 89.6 liters.
Task 5
What else are proposed in the school curriculum in chemistry for solution problems? We give an example of a task related to the need to calculate the mass of a product.
Lead (II) sulfide, having a mass of 95.6 g, interacts with 300 milliliters of a 30% hydrogen peroxide solution (density 1.1222 g / ml). The reaction product is (in grams) ...
The procedure for solving the problem:
- We translate solutions of substances through proportions to mass.
- Next, determine the amount of each source component.
- After comparing the results, we choose the substance that is taken in insufficient quantity.
- The calculations are carried out precisely for the substance taken in deficiency.
- We compose the equation of chemical interaction and calculate the mass of an unknown substance.
We calculate the peroxide solution, it is 336.66 grams. The mass of the substance will correspond to 100.99 grams. We calculate the number of moles, it will be 2.97. Lead sulfide will be 95.6 / 239 = 0.4 mol, (it is contained in a deficiency).
We make up the equation of chemical interaction. We determine the desired value according to the scheme and get 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) sulfide having a purity of 80%.
Procedure:
- Calculate the mass of pure FeS.
- We write the equation of its chemical interaction with atmospheric oxygen. We carry out calculations according to the reaction.
By mass, the pure substance will be 4488 g. The amount of the determined component will be 51 liters.
Task 7
A solution was prepared from 134.4 liters (under normal conditions) of sulfur oxide (4). 1.5 liters of a 25% sodium hydroxide solution (1.28 g / ml) were added to it. Determine the mass of the resulting salt.
Computation Algorithm:
- We calculate the mass of the alkali solution according to the formula.
- We find the mass and number of moles of caustic soda.
- We calculate the same value for sulfur oxide (4).
- By the ratio of the obtained indicators, we determine the composition of the formed salt, determine the disadvantage. Calculations are carried out at a disadvantage.
- We write down the chemical reaction with the coefficients, calculate the mass of the new salt by the deficiency.
As a result, we get:
- alkali solution 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 calculated for sulfur oxide (4), we obtain 6 mol;
- as a result of the interaction, an average salt will form;
- we get 756 grams.
Task 8
100 g of a 10% solution of silver nitrate were added to 100 grams of a 10% solution of ammonium chloride . Determine the mass (in grams) of sediment.
Computation Algorithm:
- Calculate the mass and amount of the substance of ammonium chloride.
- We calculate the mass and amount of the salt substance - silver nitrate.
- We determine which of the starting materials was taken in insufficient quantities, we carry out calculations on it.
- We write down the equation of the reaction that takes place, and we use it to calculate the mass of the precipitate.
Ammonium chloride by mass will be 10 g, by quantity - 0.19 mol. Silver nitrate is taken 10 grams, which is 0.059 mol. When calculating the deficiency, 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 course of non-boundary chemistry), you must have algorithms and have certain computing skills. In addition, it is important to own the technology of preparing chemical equations, to be able to arrange the coefficients in the process.
Without such basic 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.