Solution A homogeneous mixture of two or more components is called.
Substances that are mixed to form a solution are called components.
The components of the solution are solute, which may be more than one, and solvent. For example, in the case of a solution of sugar in water, sugar is the solute and water is the solvent.
Sometimes the concept of solvent can be applied equally to any of the components. For example, this applies to those solutions that are obtained by mixing two or more liquids that are ideally soluble in each other. So, in particular, in a solution consisting of alcohol and water, both alcohol and water can be called a solvent. However, most often in relation to water-containing solutions, it is traditionally customary to call water a solvent, and the second component the solute.
As a quantitative characteristic of the composition of the solution, such a concept is most often used as mass fraction substances in solution. The mass fraction of a substance is the ratio of the mass of this substance to the mass of the solution in which it is contained:
where ω (in-va) - mass fraction of the substance contained in the solution (g), m(v-va) - the mass of the substance contained in the solution (g), m (p-ra) - the mass of the solution (g).
From formula (1) it follows that the mass fraction can take values from 0 to 1, that is, it is a fraction of a unit. In this regard, the mass fraction can also be expressed as a percentage (%), and it is in this format that it appears in almost all problems. The mass fraction, expressed as a percentage, is calculated using a formula similar to formula (1), with the only difference that the ratio of the mass of the solute to the mass of the entire solution is multiplied by 100%:
For a solution consisting of only two components, the solute mass fraction ω(r.v.) and the solvent mass fraction ω(solvent) can be respectively calculated.
The mass fraction of a solute is also called solution concentration.
For a two-component solution, its mass is the sum of the masses of the solute and the solvent:
Also in the case of a two-component solution, the sum of the mass fractions of the solute and solvent is always 100%:
Obviously, in addition to the formulas written above, one should also know all those formulas that are directly mathematically derived from them. For example:
It is also necessary to remember the formula that relates the mass, volume and density of a substance:
m = ρ∙V
and you also need to know that the density of water is 1 g / ml. For this reason, the volume of water in milliliters is numerically equal to mass water in grams. For example, 10 ml of water has a mass of 10 g, 200 ml - 200 g, etc.
In order to successfully solve problems, in addition to knowing the above formulas, it is extremely important to bring the skills of their application to automaticity. This can only be achieved by solving a large number of different tasks. Tasks from real USE exams on the topic "Calculations using the concept of" mass fraction of a substance in solution "" can be solved.
Examples of tasks for solutions
Example 1
Calculate the mass fraction of potassium nitrate in a solution obtained by mixing 5 g of salt and 20 g of water.
Solution:
The solute in our case is potassium nitrate, and the solvent is water. Therefore, formulas (2) and (3) can be written respectively as:
From the condition m (KNO 3) \u003d 5 g, and m (H 2 O) \u003d 20 g, therefore:
Example 2
What mass of water must be added to 20 g of glucose to obtain a 10% glucose solution.
Solution:
It follows from the conditions of the problem that the solute is glucose, and the solvent is water. Then formula (4) can be written in our case as follows:
From the condition, we know the mass fraction (concentration) of glucose and the mass of glucose itself. Denoting the mass of water as x g, we can write the following equivalent equation based on the formula above:
Solving this equation we find x:
those. m(H 2 O) \u003d x g \u003d 180 g
Answer: m (H 2 O) \u003d 180 g
Example 3
150 g of a 15% sodium chloride solution were mixed with 100 g of a 20% solution of the same salt. What is the mass fraction of salt in the resulting solution? Give your answer to the nearest integer.
Solution:
To solve problems for the preparation of solutions, it is convenient to use the following table:
1st solution |
2nd solution |
3rd solution |
|
m r.v. |
|||
m solution |
|||
ω r.v. |
where m r.v. , m r-ra and ω r.v. are the values of the mass of the solute, the mass of the solution, and mass fraction solute, respectively, individual for each of the solutions.
From the condition, we know that:
m (1) solution = 150 g,
ω (1) r.v. = 15%,
m (2) solution = 100 g,
ω (1) r.v. = 20%,
Inserting all these values into the table, we get:
We should remember the following formulas required for calculations:
ω r.v. = 100% ∙ m r.v. /m solution, m r.v. = m r-ra ∙ ω r.v. / 100% , m solution = 100% ∙ m r.v. /ω r.v.
Let's start filling out the table.
If only one value is missing in a row or column, then it can be counted. The exception is the line with ω r.v., knowing the values in two of its cells, the value in the third one cannot be calculated.
The first column is missing a value in only one cell. So we can calculate it:
m (1) r.v. = m (1) r-ra ∙ ω (1) r.v. /100% = 150 g ∙ 15%/100% = 22.5 g
Similarly, we know the values in two cells of the second column, which means:
m (2) r.v. = m (2) r-ra ∙ ω (2) r.v. /100% = 100 g ∙ 20%/100% = 20 g
Let's enter the calculated values in the table:
Now we have two values in the first line and two values in the second line. So we can calculate the missing values (m (3) r.v. and m (3) r-ra):
m (3) r.v. = m (1) r.v. + m (2)r.v. = 22.5 g + 20 g = 42.5 g
m (3) solution = m (1) solution + m (2) solution = 150 g + 100 g = 250 g.
Let's enter the calculated values in the table, we get:
Now we have come close to calculating the desired value ω (3) r.v. . In the column where it is located, the contents of the other two cells are known, so we can calculate it:
ω (3)r.v. = 100% ∙ m (3) r.v. / m (3) solution = 100% ∙ 42.5 g / 250 g = 17%
Example 4
To 200 g of a 15% sodium chloride solution was added 50 ml of water. What is the mass fraction of salt in the resulting solution. Give your answer to the nearest hundredth _______%
Solution:
First of all, you should pay attention to the fact that instead of the mass of added water, we are given its volume. We calculate its mass, knowing that the density of water is 1 g / ml:
m ext. (H 2 O) = V ext. (H 2 O) ∙ ρ (H2O) = 50 ml ∙ 1 g/ml = 50 g
If we consider water as a 0% sodium chloride solution containing, respectively, 0 g of sodium chloride, the problem can be solved using the same table as in the example above. Let's draw such a table and insert the values we know into it:
In the first column, two values are known, so we can calculate the third:
m (1) r.v. = m (1)r-ra ∙ ω (1)r.v. /100% = 200 g ∙ 15%/100% = 30 g,
In the second line, two values \u200b\u200bare also known, so we can calculate the third:
m (3) solution = m (1) solution + m (2) solution = 200 g + 50 g = 250 g,
Enter the calculated values in the appropriate cells:
Now two values in the first line have become known, which means we can calculate the value of m (3) r.v. in the third cell:
m (3) r.v. = m (1) r.v. + m (2)r.v. = 30 g + 0 g = 30 g
ω (3)r.v. = 30/250 ∙ 100% = 12%.
ATTENTION!!!
STUDENTS OF 9 GRADES!!!
For successful delivery chemistry exam in some tickets you will need to solve a problem. We invite you to consider, analyze and fix in memory the solution of typical problems in chemistry.
The task is to calculate the mass fraction of a substance in solution.
50 g of phosphoric acid were dissolved in 150 g of water. Find the mass fraction of acid in the resulting solution.
Given: m(H2O) = 150g, m(H3PO4) = 50g
Find: w (H3PO4) - ?
Let's start solving the problem.
Solution: one). Find the mass of the resulting solution. To do this, simply add the mass of water and the mass of phosphoric acid added to it.
m(solution) = 150g + 50g = 200g
2). To solve, we need to know the mass fraction formula. Write down the formula for the mass fraction of a substance in a solution.
w(substances) = https://pandia.ru/text/78/038/images/image002_9.png" width="19" height="28 src="> * 100%= 25%
We write down the answer.
Answer: w(H3PO4)=25%
The task is to calculate the amount of a substance of one of the reaction products, if the mass of the initial substance is known.
Calculate the amount of iron substance that will result from the interaction of hydrogen with 480 g of iron (III) oxide.
We write down the known values in the condition of the problem.
Given: m(Fe2O3) = 4
We also write down what needs to be found as a result of solving the problem.
Find: n (Fe) - ?
Let's start solving the problem.
Solution: 1). To solve such problems, you first need to write down the reaction equation described in the problem statement.
Fe2O3 + 3 H2https://pandia.ru/text/78/038/images/image004_4.png" width="12" height="26 src="> , where n is the amount of the substance, m is the mass of this substance, and M- molar mass substances.
According to the condition of the problem, we do not know the mass of the resulting iron, i.e., in the formula for the amount of substance, we do not know two quantities. Therefore, we will look for the amount of substance by the amount of iron oxide (III) substance. The amounts of iron substance and iron(III) oxide are as follows.
https://pandia.ru/text/78/038/images/image006_4.png" height="27 src="> ; where 2 is the stoichiometric coefficient from the reaction equation in front of the iron, and 1 is the coefficient in front of the oxide iron(III).
hence n (Fe)= 2 n (Fe2O3)
3). Find the amount of iron(III) oxide substance.
n (Fe2O3) = https://pandia.ru/text/78/038/images/image008_4.png" width="43" height="20 src="> is the molar mass of iron(III) oxide, which we calculate based on the relative atomic masses of iron and oxygen, and also taking into account the number of these atoms in iron (III) oxide: M (Fe2O3) \u003d 2x 56 + 3x 16 \u003d 112 + 48 \u003d 160 aluminum rel="bookmark">aluminum ?
We write down the condition of the problem.
Given: m(Al) = 54g
And we also write down what we need to find as a result of solving the problem.
Find: V (H2) - ?
Let's start solving the problem.
Solution: 1) write the reaction equation according to the condition of the problem.
2 Al + 6 HCl https://pandia.ru/text/78/038/images/image011_1.png" width="61" height="20 src=">n - the amount of substance of a given gas.
V (H2) \u003d Vm * n (H2)
3). But in this formula, we do not know the amount of hydrogen substance.
four). Let us find the amount of hydrogen substance by the amount of aluminum substance according to the following relation.
https://pandia.ru/text/78/038/images/image013_2.png" height="27 src="> ; hence n (H2) = 3 n (Al): 2 , where 3 and 2 are stoichiometric coefficients , respectively, in front of hydrogen and aluminum.
5)..png" width="33" height="31 src=">
n (Al) = https://pandia.ru/text/78/038/images/image016_1.png" width="45" height="20 src=">* 6 mol= 134.4 L
Let's write down the answer.
Answer: V (H2) \u003d 134.4 l
The task is to calculate the amount of a substance (or volume) of a gas required for a reaction with a certain amount of a substance (or volume) of another gas.
What amount of oxygen substance is required to interact with 8 moles of hydrogen under normal conditions?
Let's write down the conditions of the problem.
Given: n (H2) = 8mol
And also write down what needs to be found as a result of solving the problem.
Find: n(O2) - ?
Let's start solving the problem.
Solution: one). We write the reaction equation following the condition of the problem.
2 H2 + O2 https://pandia.ru/text/78/038/images/image017_1.png" width="32" height="31 src="> = ; where 2 and 1 are the stoichiometric coefficients before hydrogen and oxygen, respectively, in the reaction equation.
3). Hence 2 n (O2)= n (H2)
And the amount of oxygen substance is: n (O2) \u003d n (H2): 2
four). It remains for us to substitute the data from the condition of the problem into the resulting formula.
n (O2) \u003d 8 mol: 2 \u003d 4 mol
5). Let's write down the answer.
Answer: n (O2) = 4 mol
It is known from the course of chemistry that mass fraction and name the content of a certain element in some substance. It would seem that such knowledge is of no use to an ordinary summer resident. But do not rush to close the page, as the ability to calculate the mass fraction for a gardener can be very useful. However, in order not to get confused, let's talk about everything in order.What is the meaning of the concept of "mass fraction"?
The mass fraction is measured as a percentage or simply in tenths. A little higher, we talked about the classic definition, which can be found in reference books, encyclopedias or school chemistry textbooks. But to understand the essence of what has been said is not so simple. So, suppose we have 500 g of some complex substance. Difficult in this case means that it is not homogeneous in its composition. By and large, any substances that we use are complex, even simple table salt, the formula of which is NaCl, that is, it consists of sodium and chlorine molecules. If we continue the reasoning on the example of table salt, then we can assume that 500 grams of salt contains 400 grams of sodium. Then its mass fraction will be 80% or 0.8.
Why does a gardener need this?
I think you already know the answer to this question. Preparation of all kinds of solutions, mixtures, etc. is an integral part economic activity any gardener. In the form of solutions, fertilizers, various nutrient mixtures, as well as other preparations are used, for example, growth stimulants "Epin", "Kornevin", etc. In addition, it is often necessary to mix dry substances, such as cement, sand and other components, or ordinary garden soil with purchased substrate. At the same time, the recommended concentration of these agents and preparations in prepared solutions or mixtures in most instructions is given in mass fractions.
Thus, knowing how to calculate the mass fraction of an element in a substance will help the summer resident to correctly prepare the necessary solution of fertilizer or nutrient mixture, and this, in turn, will necessarily affect the future harvest.
Calculation algorithm
So, the mass fraction of an individual component is the ratio of its mass to the total mass of a solution or substance. If the result obtained needs to be converted into a percentage, then it must be multiplied by 100. Thus, the formula for calculating the mass fraction can be written as follows:
W = Mass of substance / Mass of solution
W = (Mass of substance / Mass of solution) x 100%.
An example of determining the mass fraction
Suppose we have a solution, for the preparation of which 5 g of NaCl was added to 100 ml of water, and now it is necessary to calculate the concentration of table salt, that is, its mass fraction. We know the mass of the substance, and the mass of the resulting solution is the sum of two masses - salt and water and equals 105 g. Thus, we divide 5 g by 105 g, multiply the result by 100 and get the desired value of 4.7%. This is the concentration brine.
More practical task
In practice, the summer resident often has to deal with tasks of a different kind. For example, it is necessary to prepare an aqueous solution of a fertilizer, the concentration of which by weight should be 10%. In order to accurately observe the recommended proportions, you need to determine what amount of the substance will be needed and in what volume of water it will need to be dissolved.
Problem solving starts at reverse order. First, you should divide the mass fraction expressed as a percentage by 100. As a result, we get W \u003d 0.1 - this is the mass fraction of the substance in units. Now let's denote the amount of substance as x, and the final mass of the solution - M. In this case, the last value is made up of two terms - the mass of water and the mass of fertilizer. That is, M = Mv + x. Thus, we get a simple equation:
W = x / (Mw + x)
Solving it for x, we get:
x \u003d W x Mv / (1 - W)
Substituting the available data, we obtain the following dependence:
x \u003d 0.1 x Mv / 0.9
Thus, if we take 1 liter (that is, 1000 g) of water to prepare the solution, then approximately 111-112 g of fertilizer will be needed to prepare the solution of the desired concentration.
Solving problems with dilution or addition
Suppose we have 10 liters (10,000 g) of a ready-made aqueous solution with a concentration in it of a certain substance W1 = 30% or 0.3. How much water will need to be added to it so that the concentration drops to W2 = 15% or 0.15? In this case, the formula will help:
Mv \u003d (W1x M1 / W2) - M1
Substituting the initial data, we get that the amount of added water should be:
Mv \u003d (0.3 x 10,000 / 0.15) - 10,000 \u003d 10,000 g
That is, you need to add the same 10 liters.
Now imagine the inverse problem - there are 10 liters of an aqueous solution (M1 = 10,000 g) with a concentration of W1 = 10% or 0.1. It is necessary to obtain a solution with a mass fraction of fertilizer W2 = 20% or 0.2. How much starting material should be added? To do this, you need to use the formula:
x \u003d M1 x (W2 - W1) / (1 - W2)
Substituting the original value, we get x \u003d 1 125 g.
Thus, knowledge of the simplest basics school chemistry will help the gardener to properly prepare fertilizer solutions, nutrient substrates from several elements or mixtures for construction work.
This lesson is devoted to the study of the topic "Mass fraction of a substance in a solution." Using the lesson materials, you will learn how to quantify the content of a solute in a solution, as well as determine the composition of a solution from data on the mass fraction of a solute.
Topic: Classes of inorganic substances
Lesson: Mass fraction of a substance in solution
The mass of the solution is the sum of the masses of the solvent and the solute:
m (p) \u003d m (c) + m (r-la)
The mass fraction of a substance in a solution is equal to the ratio of the mass of the solute to the mass of the entire solution:
We will solve several problems using the above formulas.
Calculate the mass fraction (in%) of sucrose in a solution containing 250 g of water and 50 g of sucrose.
The mass fraction of sucrose in solution can be calculated using the well-known formula:
Substitute numerical values and find the mass fraction of sucrose in the solution. Received 16.7% in response.
By transforming the formula for calculating the mass fraction of a substance in a solution, you can find the values of the mass of a solute by known mass solution and mass fraction of the substance in the solution; or the mass of the solvent by the mass of the solute and the mass fraction of the substance in solution.
Consider the solution of a problem in which the mass fraction of a solute changes when the solution is diluted.
To 120 g of a solution with a mass fraction of salt of 7% was added 30 g of water. Determine the mass fraction of salt in the resulting solution.
Let's analyze the condition of the problem. In the process of diluting the solution, the mass of the solute does not change, but the mass of the solvent increases, which means that the mass of the solution increases and, conversely, the mass fraction of the substance in the solution decreases.
First, we determine the mass of the solute, knowing the mass of the initial solution and the mass fraction of salt in this solution. The mass of a solute is equal to the product of the mass of the solution and the mass fraction of the substance in the solution.
We have already found out that the mass of a solute does not change when a solution is diluted. This means that by calculating the mass of the resulting solution, you can find the mass fraction of salt in the resulting solution.
The mass of the resulting solution is equal to the sum of the masses of the initial solution and the added water. The mass fraction of salt in the resulting solution is equal to the ratio of the mass of the solute and the mass of the resulting solution. Thus, a mass fraction of salt in the resulting solution was obtained equal to 5.6%.
1. Collection of tasks and exercises in chemistry: 8th grade: to textbook. P.A. Orzhekovsky and others. “Chemistry. Grade 8 / P.A. Orzhekovsky, N.A. Titov, F.F. Hegel. - M .: AST: Astrel, 2006. (p. 111-116)
2. Ushakova O.V. Chemistry workbook: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry. Grade 8” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; ed. prof. P.A. Orzhekovsky - M .: AST: Astrel: Profizdat, 2006. (p. 111-115)
3. Chemistry. 8th grade. Textbook for general education institutions / P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova. – M.: Astrel, 2013. (§35)
4. Chemistry: 8th grade: textbook for general education. institutions / P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. M.: AST: Astrel, 2005. (§41)
5. Chemistry: inorg. chemistry: textbook for 8 cells. general education institutions / G.E. Rudzitis, F.G. Feldman. - M .: Education, JSC "Moscow textbooks", 2009. (§ 28)
6. Encyclopedia for children. Volume 17. Chemistry / Editor-in-Chief. V.A. Volodin, leading. scientific ed. I. Leenson. – M.: Avanta+, 2003.
Additional web resources
3. Interaction of substances with water ().
Homework
1. p. 113-114 №№ 9,10 from Workbook in chemistry: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry. Grade 8” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; ed. prof. P.A. Orzhekovsky - M.: AST: Astrel: Profizdat, 2006.
2. p.197 №№ 1,2 from the textbook P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova "Chemistry: 8th grade", 2013
