ATTENTION!!!
STUDENTS OF 9 CLASSES !!!
For successful delivery chemistry exam in some tickets you will need to solve a problem. We invite you to consider, disassemble and fix in your memory the solution of typical problems in chemistry.
The task of calculating the mass fraction of a substance in solution.
In 150 g of water, 50 g of phosphoric acid was dissolved. 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: 1). We 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. We 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 of calculating 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 that will result from the interaction of hydrogen with 480 g of iron (III) oxide.
We write the known values into 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 H2 M - molar mass substances.
By the condition of the problem, we do not know the mass of the resulting iron, that is, in the formula for the amount of matter, we do not know two quantities. Therefore, we will look for the amount of a substance by the amount of an iron (III) oxide substance. The amount of iron substance and iron oxide (III) by the following ratio.
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 iron, and 1 is the coefficient in front of 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, as well as taking into account the number of these atoms in iron (III) oxide: М (Fe2O3) = 2х 56 + 3х 16 = 112 + 48 = 160 Aluminum "href =" / text / category / alyuminij / " 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) we write down 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 is the amount of substance of a given gas.
V (H2) = Vm * n (H2)
3). But in this formula we do not know the amount of hydrogen substance.
4). Let us find the amount of hydrogen substance by the amount of aluminum substance according to the following ratio.
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 facing hydrogen and aluminum, respectively.
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) = 134.4 l
The task of calculating the amount of a substance (or volume) of a gas required to react with a certain amount of a substance (or volume) of another gas.
How much oxygen is required to interact with 8 moles of hydrogen under normal conditions?
Let's write down the conditions of the problem.
Given: n (H2) = 8 mol
And we will 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: 1). Let us write down the reaction equation following the condition of the problem.
2 H2 + О2https: //pandia.ru/text/78/038/images/image017_1.png "width =" 32 "height =" 31 src = "> =; where 2 and 1 are 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) = n (H2): 2
4). It remains for us to substitute the data from the problem statement into the resulting formula.
n (О2) = 8 mol: 2 = 4 mol
5). Let's write down the answer.
Answer: n (О2) = 4 mol
Solution refers to a homogeneous mixture of two or more components.
The substances, by mixing which the solution is obtained, call it components.
Among the components of the solution are distinguished solute which may not be one, and solvent... For example, in the case of a solution of sugar in water, sugar is a solute and water is a solvent.
Sometimes the concept of a 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, 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 aqueous solutions, it is customary to call water a solvent, and a second component as a dissolved substance.
As a quantitative characteristic of the composition of the solution, the most often used concept is 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(in-va) - the mass of the substance contained in the solution (g), m (solution) - 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 mass fraction of the solute ω (r.v.) and the mass fraction of the solvent ω (solvent) can be calculated accordingly.
The mass fraction of the solute is also called concentration of solution.
For a two-component solution, its mass consists 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 the solvent is always 100%:
Obviously, in addition to the formulas written above, you should know all those formulas that are mathematically derived directly 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 automatism. This can be achieved only by solving a large number of different problems. Tasks from real examinations on the topic "Calculations using the concept of" mass fraction of a substance in a solution "" can be resolved.
Examples of problems 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 dissolved substance 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) = 5 g, and m (H 2 O) = 20 g, therefore:
Example 2
What mass of water must be added to 20 g of glucose to obtain a 10% glucose solution.
Solution:
From the conditions of the problem it follows 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 down the following equivalent equation based on the formula above:
Solving this equation, we find x:
those. m (H 2 O) = x g = 180 g
Answer: m (H 2 O) = 180 g
Example 3
150 g of a 15% sodium chloride solution was mixed with 100 g of a 20% solution of the same salt. What is the mass fraction of salt in the resulting solution? Indicate your answer to the nearest whole.
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 solution and ω r.v. - the values of the mass of the solute, the mass of the solution and the mass fraction of the 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%,
Let's insert 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 solution ∙ ω r.v. / 100%, m solution = 100% ∙ m r.v. / ω r.v.
We begin to fill in the table.
If only one value is missing in a row or column, then it can be calculated. An exception is a line with ω r.v., knowing the values in two of its cells, the value in the third 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 into the table:
Now we know 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 into the table, we get:
Now we have come close to calculating the required value of ω (3) r.v. ... In the column where it is located, the contents of the other two cells are known, which means 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. Indicate 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. Let's calculate its mass, knowing that the density of water is 1 g / ml:
m ext. (H 2 O) = V ext. (H 2 O) ∙ ρ (H 2 O) = 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, which means 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 are also known, which means we can calculate the third:
m (3) solution = m (1) solution + m (2) solution = 200 g + 50 g = 250 g,
Let's enter the calculated values into the corresponding 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%.
1. Fill in the blanks.
a) Solution = solute+ solvent;
b) m (solution) = m (solute)+ m (solvent).
2. Make a definition using the following words:
mass fraction, substance, mass, solution, to mass, ratio, in solution, substance, dissolved.
Answer: the mass fraction of a substance in a solution is the ratio of the mass of the solute to the mass of the solution.
3. Compose the formulas using the welchin notation.
| m | m solution | V |
| p = m / V | w = m (substance) / m (solution) | m = w * m (solution) |
4. What is the mass fraction of a solute if it is known that 80 g of a solution contains 20 g of salt?
5. Determine the masses of salt and water that will be required to prepare 300 g of a solution with a mass fraction of 20% salt.
6. Calculate the mass of water required to prepare 60 g of 10% salt solution.
7. The pharmacy sells powder "Regidron", which is used for dehydration. One packet of powder contains 3–5 g of sodium chloride, 2–5 g of potassium chloride, 2–9 g of sodium citrate and 10 g of glucose. The contents of the packet are dissolved in 1 liter of water. Determine the mass fractions of all components of the "Regidron" powder in the resulting solution.
8. To 500 g of 20% glucose solution was added 300 g of water. Calculate the mass fraction of glucose in the new solution.
9. To 400 g of 5% sodium chloride solution was added 50 g of salt. Calculate the mass fraction of sodium chloride in the new solution.
10. Merged two salt solutions: 100 g of 20% and 450 g of 10%. Calculate the mass fraction of salt in the new solution.
Instructions
Mass fraction is the ratio of the mass of the solute to the mass solution... Moreover, it can be measured or, then for this, the result obtained must be multiplied by 100% or in mass fractions (in this case, it does not have units).
Any solution consists of (water is the most common solvent) and a solute. For example, in any salt solution, the solvent will be water, and the salt itself will act as a solute.
For calculations, you need to know at least two parameters - the mass of water and the mass of salt. This will make it possible to calculate the mass share substance which is w (omega).
Example 1. Weight solution hydroxide (KOH) 150 g, mass of solute (KOH) 20 g. Find the mass share(KOH) in the resulting solution.
m (KOH) = 20 g
m (KOH) = 100 g
w (KOH) -? Exists, by which it is possible to determine the mass share substances.
w (KOH) = m (KOH) / m ( solution(KOH) x 100% Now calculate the mass share solute potassium hydroxide (KOH):
w (KOH) = 20 g / 120 g x 100% = 16.6%
Example 2. The mass of water is 100 g, the mass of salt is 20 g. Find the mass share chloride in solution.
m (NaCl) = 20 g
m (water) = 100 g
w (NaCl) -? There is a formula by which you can determine the mass share substances.
w (NaCl) = m (NaCl) / m ( solution NaCl) x 100% Before using this formula, find the mass solution, which consists of the mass of the solute and the mass of water. Therefore: m ( solution NaCl) = m (solute NaCl) + m (water) Substitute specific values
m ( solution NaCl) = 100 g + 20 g = 120 g Now calculate the mass share solute:
w (NaCl) = 20 g / 120 g x 100% = 16.7%
When calculating, do not confuse concepts such as mass of solute and mass fraction of solute.
The mass fraction of a substance shows its content in a more complex structure, for example, in an alloy or mixture. If the total mass of the mixture or alloy is known, then knowing the mass fractions of the constituent substances, you can find their masses. You can find the mass fraction of a substance by knowing its mass and the mass of the entire mixture. This value can be expressed in fractions or percentages.
You will need
- scales;
- periodic table of chemical elements;
- calculator.
Instructions
Determine the mass fraction of the substance that is in the mixture through the mass of the mixture and the substance itself. To do this, using the balance, determine the masses that make up the mixture or. Then fold them up. Take the resulting mass as 100%. To find the mass fraction of a substance in a mixture, divide its mass m by the mass of the mixture M, and multiply the result by 100% (ω% = (m / M) ∙ 100%). For example, 20 g of sodium chloride is dissolved in 140 g of water. To find the mass fraction of salt, add the masses of these two substances M = 140 + 20 = 160 g. Then find the mass fraction of the substance ω% = (20/160) ∙ 100% = 12.5%.
