A task.

Calculate the mass of salt and water required to prepare 40 g of NaCl solution with a mass fraction of 5%.

1. Write down the condition of the problem using generally accepted notation

m solution = 40g

1. Calculate the mass of the solute using the formula:

m in-va \u003d ω ∙ m solution / 100%

m (NaCl) \u003d 5% 40g / 100% \u003d 2g

2. Find the mass of water by the difference between the mass of the solution and the mass of the solute:

m r-la \u003d m r-ra - m in-va

m (H 2 O) \u003d 40g - 2g \u003d 38 g.

3. Write down the answer.

Answer: to prepare the solution, you need to take 2g of salt and 38g of water.

Algorithm for finding the mass fraction of a solute when diluting (evaporating) a solution

A task

m r-ra1 = 80g

m (H 2 O) \u003d 30g

1. As a result of dilution (evaporation) of the solution, the mass of the solution increased (decreased), and the same amount of substance remained in it.

Calculate the mass of the solute by converting the formula:

ω \u003d m in-va / m solution ∙ 100%

m in-va \u003d ω 1 m r-ra1 / 100%

m in-va \u003d 15% 80g \u003d 12g

2. When the solution is diluted, its total mass increases (when evaporated, it decreases).

Find the mass of the newly obtained solution:

m r-ra2 \u003d m r-ra1 + m (H 2 O)

m solution2 = 80g + 30g = 110g

3. Calculate the mass fraction of the solute in the new solution:

ω 2 \u003d m in-va / m r-ra2 ∙ 100%

ω 2 \u003d 12g / 110g 100% \u003d 10.9%

4. Write down the answer

Answer: the mass fraction of the solute in the solution when diluted is 10.9%

Algorithm for solving problems according to the "rule of the cross"

To obtain a solution with a given mass fraction (%) of a solute by mixing two solutions with a known mass fraction of a solute, a diagonal scheme is used ("rule of the cross").

The essence of this method is that the smaller one is subtracted diagonally from the larger value of the mass fraction of the solute.

The differences (s-b) and (a-c) show the ratios in which solutions a and b must be taken in order to obtain solution c.

If a pure solvent, for example, H 2 0, is used for dilution as the initial solution, then its concentration is taken as 0 and is written on the left side of the diagonal diagram.

A task

For the treatment of the surgeon's hands, wounds, postoperative field, iodine tincture with a mass fraction of 5% is used. In what mass ratio should solutions with mass fractions of iodine 2.5% and 30% be mixed in order to obtain 330 g of iodine tincture with a mass fraction of iodine 5%?

1. Write down the condition of the problem using generally accepted notation.

1. Make a "diagonal pattern". To do this, write down the mass fractions of the initial solutions one below the other, on the left side of the cross, and in the center the given mass fraction of the solution.

2. Subtract from the larger mass fraction the smaller one (30–5=25; 5–2.5=2.5) and find the results.

Write down the results found on the right side of the diagonal scheme: if possible, reduce the numbers obtained. In this case, 25 is ten times greater than 2.5, that is, instead of 25 they write 10, instead of 2.5 they write 1.

The numbers (in this case 25 and 2.5 or 10 and 1) are called mass numbers. Mass numbers show in what ratio it is necessary to take the initial solutions in order to obtain a solution with a mass fraction of iodine of 5%.

3. Determine the mass of 30% and 2.5% solution using the formula:

m p-ra \u003d number of parts m 3 / sum of mass parts

m 1 (30%) \u003d 1 330g / 1 + 10 \u003d 30g

m 2 (2.5%) \u003d 10 330g / 1 + 10 \u003d 300g

4. Write down the answer.

Answer: to prepare 330 g of a solution with a mass fraction of iodine of 5%, it is necessary to mix 300 g of a solution with a mass fraction of 2.5% and 30 g with a mass fraction of 30%.

Task 3.1. Determine the mass of water in 250 g of a 10% sodium chloride solution.

Solution. From w \u003d m in-va / m solution find the mass of sodium chloride:
m in-va \u003d w m solution \u003d 0.1 250 g \u003d 25 g NaCl
Insofar as m r-ra = m in-va + m r-la, then we get:
m (H 2 0) \u003d m solution - m in-va \u003d 250 g - 25 g \u003d 225 g H 2 0.

Task 3.2. Determine the mass of hydrogen chloride in 400 ml of solution of hydrochloric acid with a mass fraction of 0.262 and a density of 1.13 g / ml.

Solution. Insofar as w = m in-va / (V ρ), then we get:
m in-va \u003d w V ρ \u003d 0.262 400 ml 1.13 g / ml \u003d 118 g

Task 3.3. To 200 g of a 14% salt solution was added 80 g of water. Determine the mass fraction of salt in the resulting solution.

Solution. Find the mass of salt in the original solution:
m salt \u003d w m solution \u003d 0.14 200 g \u003d 28 g.
The same mass of salt remained in the new solution. Find the mass of the new solution:
m solution = 200 g + 80 g = 280 g.
Find the mass fraction of salt in the resulting solution:
w \u003d m salt / m solution \u003d 28 g / 280 g \u003d 0.100.

Task 3.4. What volume of a 78% sulfuric acid solution with a density of 1.70 g/ml should be taken to prepare 500 ml of a 12% sulfuric acid solution with a density of 1.08 g/ml?

Solution. For the first solution we have:
w 1 \u003d 0.78 And ρ 1 \u003d 1.70 g / ml.
For the second solution we have:
V 2 \u003d 500 ml, w 2 \u003d 0.12 And ρ 2 \u003d 1.08 g / ml.
Since the second solution is prepared from the first by adding water, the masses of the substance in both solutions are the same. Find the mass of the substance in the second solution. From w 2 \u003d m 2 / (V 2 ρ 2) we have:
m 2 \u003d w 2 V 2 ρ 2 \u003d 0.12 500 ml 1.08 g / ml \u003d 64.8 g.
m 2 \u003d 64.8 g. We find
the volume of the first solution. From w 1 = m 1 / (V 1 ρ 1) we have:
V 1 \u003d m 1 / (w 1 ρ 1) \u003d 64.8 g / (0.78 1.70 g / ml) \u003d 48.9 ml.

Task 3.5. What volume of a 4.65% sodium hydroxide solution with a density of 1.05 g/ml can be prepared from 50 ml of a 30% sodium hydroxide solution with a density of 1.33 g/ml?

Solution. For the first solution we have:
w 1 \u003d 0.0465 And ρ 1 \u003d 1.05 g / ml.
For the second solution we have:
V 2 \u003d 50 ml, w 2 \u003d 0.30 And ρ 2 \u003d 1.33 g / ml.
Since the first solution is prepared from the second by adding water, the masses of the substance in both solutions are the same. Find the mass of the substance in the second solution. From w 2 \u003d m 2 / (V 2 ρ 2) we have:
m 2 \u003d w 2 V 2 ρ 2 \u003d 0.30 50 ml 1.33 g / ml \u003d 19.95 g.
The mass of the substance in the first solution is also equal to m 2 \u003d 19.95 g.
Find the volume of the first solution. From w 1 = m 1 / (V 1 ρ 1) we have:
V 1 \u003d m 1 / (w 1 ρ 1) \u003d 19.95 g / (0.0465 1.05 g / ml) \u003d 409 ml.
Solubility coefficient (solubility) - the maximum mass of a substance soluble in 100 g of water at a given temperature. A saturated solution is a solution of a substance that is in equilibrium with the existing precipitate of that substance.

Problem 3.6. The solubility coefficient of potassium chlorate at 25 ° C is 8.6 g. Determine the mass fraction of this salt in saturated solution at 25 °C.

Solution. 8.6 g of salt dissolved in 100 g of water.
The mass of the solution is:
m solution \u003d m water + m salt \u003d 100 g + 8.6 g \u003d 108.6 g,
and the mass fraction of salt in the solution is equal to:
w \u003d m salt / m solution \u003d 8.6 g / 108.6 g \u003d 0.0792.

Problem 3.7. The mass fraction of salt in a potassium chloride solution saturated at 20 °C is 0.256. Determine the solubility of this salt in 100 g of water.

Solution. Let the solubility of the salt be X g in 100 g of water.
Then the mass of the solution is:
m solution = m water + m salt = (x + 100) g,
and the mass fraction is:
w \u003d m salt / m solution \u003d x / (100 + x) \u003d 0.256.
From here
x = 25.6 + 0.256x; 0.744x = 25.6; x = 34.4 g per 100 g of water.
Molar concentration from- the ratio of the amount of solute v (mol) to the volume of the solution V (in liters), c \u003d v (mol) / V (l), c \u003d m in-va / (M V (l)).
Molar concentration shows the number of moles of a substance in 1 liter of solution: if the solution is decimolar ( c = 0.1 M = 0.1 mol/l) means that 1 liter of the solution contains 0.1 mol of the substance.

Problem 3.8. Determine the mass of KOH required to prepare 4 liters of a 2 M solution.

Solution. For solutions with a molar concentration, we have:
c \u003d m / (M V),
where from- molar concentration,
m- the mass of the substance,
M is the molar mass of the substance,
V- the volume of the solution in liters.
From here
m \u003d c M V (l) \u003d 2 mol / l 56 g / mol 4 l \u003d 448 g KOH.

Problem 3.9. How many ml of a 98% solution of H 2 SO 4 (ρ = 1.84 g / ml) must be taken to prepare 1500 ml of a 0.25 M solution?

Solution. The task of diluting the solution. For a concentrated solution we have:
w 1 \u003d m 1 / (V 1 (ml) ρ 1).
Find the volume of this solution V 1 (ml) \u003d m 1 / (w 1 ρ 1).
Since a dilute solution is prepared from a concentrated solution by mixing the latter with water, the mass of the substance in these two solutions will be the same.
For a dilute solution we have:
c 2 \u003d m 2 / (M V 2 (l)) And m 2 \u003d s 2 M V 2 (l).
We substitute the found value of the mass into the expression for the volume of a concentrated solution and carry out the necessary calculations:
V 1 (ml) \u003d m / (w 1 ρ 1) \u003d (s 2 M V 2) / (w 1 ρ 1) \u003d (0.25 mol / l 98 g / mol 1.5 l) / (0, 98 1.84 g/ml) = 20.4 ml.

Concentration calculations
solutes
in solutions

Solving the problems of diluting solutions is not particularly difficult, but requires care and some tension. Nevertheless, it is possible to simplify the solution of these problems by using the dilution law, which is used in analytical chemistry when titrating solutions.
All problem books in chemistry show solutions to problems presented as a sample solution, and all solutions use the law of dilution, the principle of which is that the amount of a solute and the mass m remain unchanged in the original and diluted solutions. When we solve a problem, we keep this condition in mind, and we write down the calculation in parts and gradually, step by step, we approach the final result.
Consider the problem of solving dilution problems based on the following considerations.

Amount of solute:

= c V,

where c is the molar concentration of the solute in mol/l, V- the volume of the solution in l.

Mass of solute m(r.v.):

m(r.v.) = m(r-ra),

where m(p-ra) - the mass of the solution in g, - the mass fraction of the solute.
Let us denote in the initial (or undiluted) solution the quantities c, V, m(r-ra), through from 1 ,V 1 ,
m 1 (p-ra), 1, and in a dilute solution - through from 2 ,V 2 ,m 2 (r-ra), 2 .
Let's make the equations of dilution of solutions. The left parts of the equations will be taken for the initial (undiluted) solutions, and the right parts - for dilute solutions.
The constancy of the amount of a solute upon dilution will have the form:

Conservation of mass m(r.v.):

The amount of a solute is related to its mass m(r.v.) ratio:

= m(r.v.) / M(r.v.),

where M(r.v.) is the molar mass of the solute in g/mol.
Dilution equations (1) and (2) are interconnected as follows:

from 1 V 1 = m 2 (r-ra) 2 / M(r.v.),

m 1 (r-ra) 1 = from 2 V 2 M(R.V.).

If the volume of dissolved gas is known in the problem V(gas), then its amount of substance is related to the volume of gas (n.o.) by the ratio:

= V(gas)/22.4.

The dilution equations will take the form, respectively:

V(gas)/22.4 = from 2 V 2 ,

V(gas)/22.4 = m 2 (r-ra) 2 / M(gas).

If the mass of a substance or the amount of a substance taken to prepare a solution are known in the problem, then on the left side of the dilution equation is put m(r.v.) or , depending on the condition of the problem.
If, according to the condition of the problem, it is required to combine solutions of different concentrations of the same substance, then the masses of the dissolved substances are summed on the left side of the equation.
Quite often, in tasks, the density of the solution (g / ml) is used. But since the molar concentration from measured in mol / l, then the density should be expressed in g / l, and the volume V- in l.
Let us give examples of solving "exemplary" problems.

Task 1. What volume of 1M sulfuric acid solution should be taken to obtain 0.5 l of 0.1M H2SO4 ?

Given:

c 1 \u003d 1 mol / l,
V 2 = 0.5 l,
from 2 = 0.1 mol/l.

To find:

Solution

V 1 from 1 =V 2 from 2 ,

V 1 1 \u003d 0.5 0.1; V 1 = 0.05 l, or 50 ml.

Answer.V 1 = 50 ml.

Task 2 (, № 4.23). Determine the mass of the solution with a mass fraction(CuSO 4) 10% and the mass of water that will be required to prepare a solution weighing 500 g with a mass fraction
(CuSO 4) 2%.

Given:

1 = 0,1,
m 2 (r-ra) = 500 g,
2 = 0,02.

To find:

m 1 (r-ra) =?
m(H 2 O) \u003d?

Solution

m 1 (r-ra) 1 = m 2 (r-ra) 2,

m 1 (r-ra) 0.1 \u003d 500 0.02.

From here m 1 (r-ra) \u003d 100 g.

Find the mass of added water:

m(H 2 O) = m 2 (r-ra) - m 1 (r-ra),

m (H 2 O) \u003d 500 - 100 \u003d 400 g.

Answer. m 1 (r-ra) \u003d 100 g, m(H 2 O) \u003d 400 g.

Task 3 (, № 4.37).What is the volume of a solution with a mass fraction of sulfuric acid 9.3%
(\u003d 1.05 g / ml) will be required to prepare 0.35 M solution H2SO4 40 ml?

Given:

1 = 0,093,
1 = 1050 g/l,
from 2 = 0.35 mol/l,
V 2 = 0.04 l,
M(H 2 SO 4) \u003d 98 g / mol.

To find:

Solution

m 1 (r-ra) 1 = V 2 from 2 M(H 2 SO 4),

V 1 1 1 = V 2 from 2 M(H2SO4).

We substitute the values ​​of known quantities:

V 1 1050 0.093 = 0.04 0.35 98.

From here V 1 \u003d 0.01405 l, or 14.05 ml.

Answer. V 1 = 14.05 ml.

Task 4 . What volume of hydrogen chloride (n.o.) and water will be required to prepare 1 liter of solution (\u003d 1.05 g / cm 3), in which the content of hydrogen chloride in mass fractions is 0.1
(or 10%)?

Given:

V (solution) \u003d 1 l,
(solution) = 1050 g/l,
= 0,1,
M(HCl) = 36.5 g/mol.

To find:

V(HCl) = ?
m(H 2 O) \u003d?

Solution

V(HCl)/22.4 = m(p-ra) / M(HCl)

V(HCl)/22.4 = V(r-ra) (r-ra) / M(HCl)

V(HCl)/22.4 = 1 1050 0.1/36.5.

From here V(HCl) = 64.44 liters.
Find the mass of added water:

m(H 2 O) = m(r-ra) - m(HCl),

m(H 2 O) = V(r-ra) (r-ra) - V(HCl)/22.4 M(HCl)

m (H 2 O) \u003d 1 1050 - 64.44 / 22.4 36.5 \u003d 945 g.

Answer. 64.44 L HCl and 945 g water.

Task 5 (, № 4.34). Determine molar concentration solution with a mass fraction of sodium hydroxide 0.2 and a density of 1.22 g / ml.

Given:

0,2,
= 1220 g/l,
M(NaOH) = 40 g/mol.

To find:

Solution

m(p-ra) = from V M(NaOH),

m(p-ra) = from m(r-ra) M(NaOH)/.

Divide both sides of the equation by m(r-ra) and substitute numerical values quantities.

0,2 = c 40/1220.

From here c= 6.1 mol/l.

Answer. c= 6.1 mol/l.

Task 6 (, № 4.30).Determine the molar concentration of the solution obtained by dissolving sodium sulfate weighing 42.6 g in water weighing 300 g, if the density of the resulting solution is 1.12 g/ml.

Given:

m (Na 2 SO 4) \u003d 42.6 g,
m(H 2 O) \u003d 300 g,
= 1120 g/l,
M(Na 2 SO 4) \u003d 142 g / mol.

To find:

Solution

m(Na 2 SO 4) = from V M(Na2SO4).

500 (1 – 4,5/(4,5 + 100)) = m 1 (r-ra) (1 - 4.1 / (4.1 + 100)).

From here m 1 (solution) \u003d 104.1 / 104.5 500 \u003d 498.09 g,

m(NaF) = 500 - 498.09 = 1.91 g.

Answer. m(NaF) = 1.91 g.

LITERATURE

1.Khomchenko G.P., Khomchenko I.G. Tasks in chemistry for university students. M.: New wave, 2002.
2. Feldman F.G., Rudzitis G.E. Chemistry-9. M.: Enlightenment, 1990, p. 166.

Task 3.
5 g table salt(NaCl) was dissolved in some water. As a result, a 4% solution of NaCl in water was obtained. Determine the amount of water used.
Given:
mass of table salt: mNaCl) = 5 g;
mass fraction of NaCl in the resulting solution: NaCl) = 4%.
To find:
mass of water used.
Solution:
This problem can be solved in two ways: using a formula and proportion.

I way:

We substitute the data from the condition in the first formula and find the mass of the solution.

II way:

Schematically, the solution algorithm can be represented as follows:

The mass fraction of water in the solution is: 100% - 4% = 96%.

Since the solution contains 5 g of salt, which make up 4%, you can make up the proportion:
5 g is 4%
x g make up 96%

Answer: m of water = 120g.

Task 4.
A certain amount of pure sulfuric acid was dissolved in 70 g of water. The result was a 10% solution of H 2 SO 4 . Determine the mass of sulfuric acid used.
Given:
mass of water: m (H 2 O) \u003d 70 g;
mass fraction of H 2 SO 4 in the resulting solution: H 2 SO 4) = 10%.
To find:
mass of sulfuric acid used.
Solution:
It is also possible to use both ratio and proportion.

I way:

Let us substitute the last expression into the ratio for the mass fraction:

We substitute the data from the condition into the resulting formula:

We got one equation with one unknown Solving it, we find the mass of sulfuric acid used:

II way:

Schematically, the solution algorithm can be represented as follows:

Let's apply the proposed algorithm.

m (H 2 O) \u003d 100% - (H 2 SO 4) \u003d 100% - 10% \u003d 90%

We make a proportion:
70g make up 90%
x g make up 10%

Answer: m (H 2 SO 4) \u003d 7.8 g.

Task 5.
Some sugar was dissolved in water. As a result, 2 liters of a 30% solution were obtained (p = 1.127 g/ml). Determine the mass of dissolved sugar and the volume of water used.
Given:
solution volume: V solution = 2 l;
mass fraction of sugar in solution: (sugar) = 30%;
solution density: R solution = 1.127 g / ml
To find:
mass of dissolved sugar; volume of water used.
Solution:
Schematically, the solution algorithm can be represented as follows.

Calculation of the mass of a solution of a certain concentration from the mass of a solute or solvent.

Calculation of the mass of a solute or solvent from the mass of the solution and its concentration.

Calculation of the mass fraction (in percent) of the dissolved substance.

Examples of typical tasks for calculating the mass fraction (in percent) of a dissolved substance.

Percent concentration.

Mass fraction (in percent) or percentage concentration (ω) - shows the number of grams of a solute contained in 100 grams of a solution.

The percentage concentration or mass fraction is the ratio of the mass of the solute to the mass of the solution.

ω = msolv. in-va · 100% (1),

m solution

where ω is the percentage concentration (%),

m sol. in-va - the mass of the dissolved substance (g),

m solution - mass of solution (g).

The mass fraction is measured in fractions of a unit and is used in intermediate calculations. If the mass fraction is multiplied by 100%, the percentage concentration is obtained, which is used when the final result is issued.

The mass of the solution is the sum of the mass of the solute and the mass of the solvent:

m r-ra = m r-la + m sol. in-va (2),

where m r-ra is the mass of the solution (g),

m p-la - the mass of the solvent (g),

m sol. in-va - the mass of the dissolved substance (g).

For example, if the mass fraction of the dissolved substance - sulfuric acid in water is 0.05, then the percentage concentration is 5%. This means that a 100 g solution of sulfuric acid contains sulphuric acid weighing 5 g, and the mass of the solvent is 95 g.

EXAMPLE 1 . Calculate the percentage of crystalline hydrate and anhydrous salt if 50 g of CuSO 4 5H 2 O were dissolved in 450 g of water.

SOLUTION:

1) The total mass of the solution is 450 + 50 = 500 g.

2) The percentage of crystalline hydrate is found by the formula (1):

X \u003d 50 100 / 500 \u003d 10%

3) Calculate the mass of anhydrous CuSO 4 salt contained in 50 g of crystalline hydrate:

4) Calculate the molar mass of CuSO 4 5H 2 O and anhydrous CuSO 4

M CuSO4 5H2O = M Cu + M s +4M o + 5M H2O = 64 + 32 + 4 16 + 5 18 = 250 g/mol

M CuSO4 \u003d M Cu + M s + 4M o \u003d 64 + 32 + 4 16 \u003d 160 g / mol

5) 250 g CuSO 4 5H 2 O contains 160 g CuSO 4

And in 50 g CuSO 4 5H 2 O - X g CuSO 4

X \u003d 50 160 / 250 \u003d 32 g.

6) The percentage of anhydrous salt of copper sulfate will be:

ω = 32 100 / 500 = 6.4%

ANSWER : ω CuSO4 5H2O = 10%, ω CuSO4 = 6.4%.

EXAMPLE 2 . How many grams of salt and water are contained in 800 g of 12% NaNO 3 solution?

SOLUTION:

1) Find the mass of the solute in 800 g of a 12% solution of NaNO 3:

800 12 / 100 = 96 g

2) The mass of the solvent will be: 800 -96 \u003d 704 g.

ANSWER: Mass HNO 3 \u003d 96 g, mass H 2 O \u003d 704 g.

EXAMPLE 3 . How many grams of a 3% solution of MgSO 4 can be prepared from 100 g of MgSO 4 7H 2 O?

SOLUTION :

1) Calculate the molar mass of MgSO 4 7H 2 O and MgSO 4

M MgSO4 7H2O = 24 + 32 + 4 16 + 7 18 = 246 g/mol

M MgSO4 = 24 + 32 + 4 16 = 120 g/mol

2) 246 g MgSO 4 7H 2 O contains 120 g MgSO 4

100 g MgSO 4 7H 2 O contains X g MgSO 4

X \u003d 100 120 / 246 \u003d 48.78 g

3) According to the condition of the problem, the mass of anhydrous salt is 3%. From here:

3% of the mass of the solution is 48.78 g

100% of the mass of the solution is X g

X \u003d 100 48.78 / 3 \u003d 1626 g

ANSWER : the mass of the prepared solution will be 1626 grams.

EXAMPLE 4. How many grams of HC1 must be dissolved in 250 g of water to obtain a 10% solution of HC1?

SOLUTION: 250 g of water make up 100 - 10 \u003d 90% of the mass of the solution, then the mass of HC1 is 250 10 / 90 = 27.7 g of HC1.

ANSWER : The mass of HCl is 27.7 g.