The most intriguing area of ​​application of cryobiology - the science of the influence of low and ultra-low temperatures on biological objects - is the search for possibilities of preserving living organisms or individual organs in a state of deep freezing. The technique of cryopreservation of individual cells or, for example, embryos has been developed well, but reversible (i.e., maintaining viability after thawing) freezing of large objects encounters serious obstacles. The main difficulty is that with large volume and mass it is difficult to achieve uniform cooling. Uneven freezing leads to serious and irreversible damage to cells and tissues. Meanwhile, solving this problem could help, for example, create a bank of organs for transplantation and thereby save the lives of thousands of patients. Even more tempting is the possibility of keeping a seriously ill patient in a state of deep cooling until medicine is able to help him, perhaps decades from now.

The greatest danger during freezing is mechanical damage to cell membranes from the resulting ice crystals. Forming both outside and, which is much more dangerous, inside cells, they rupture the lipid bimolecular layer that forms these membranes.

To protect cells from damage during freezing, special substances are used - cryoprotectors. They are divided into two groups: cell-penetrating, or endocellular (dimethyl sulfoxide (DMSO), acetamide, propylene glycol, glycerol, ethylene glycol), and non-penetrating or exocellular (polyethylene glycols and polyethylene oxides, Ficoll, sucrose, trehalose, etc.), which act externally. , osmotically drawing water out of the cell.

The latter is beneficial: the less water remains in the cell, the less ice will form later. But the removal of water leads to an increase in the concentration of salts remaining inside the cell - up to values ​​​​at which denaturation of the protein occurs. Endocellular cryoprotectors not only reduce the freezing point, but also dilute the “brine” formed during crystallization, preventing proteins from denaturing.

The most widely used are glycerin and DMSO. When they are added to water, its freezing point decreases, reaching its lowest value at a ratio of approximately 2:1. This lowest temperature is called eutectic, or cryohydrate. With further cooling of such mixtures, the sizes of the resulting ice crystals turn out to be so small (comparable to the size of the crystal cell) that they do not cause significant damage to the cell structures.

If it were possible to bring the cryoprotectant concentration in living tissues to eutectic, this would completely solve the problem of tissue damage from ice crystals. However, at such concentrations, any known cryoprotectors turn out to be toxic.

In practice, cryoprotectant concentrations are used that are significantly lower than eutectic concentrations, and yet some of the water still freezes. So, when using a 27% solution of glycerol, 40% of the water present in the cell forms a eutectic mixture with glycerol, while the rest freezes. However, as experiments conducted in 1954–1960 showed. English cryobiologist Audrey Smith, golden hamsters are able to survive in a situation where up to 50-60% of the water contained in the tissues of their brain has turned into ice!

The cooling rate is of great importance for solving the problem of reversible freezing. During slow cooling (in liquid nitrogen vapor or in special program freezers), ice crystals form mainly in the intercellular space. As they cool, they grow, drawing water from the cells. As already mentioned, this can significantly reduce the damage caused by crystals to cells - but the concentration of salts inside the cells increases significantly, increasing the risk of protein denaturation.

Unfortunately, the optimal rates of temperature reduction that compromise between the damaging effects of ice crystals and high solute concentrations vary greatly among cell types. The optimal concentrations of cryoprotectants for them are also different. This greatly complicates the cryopreservation of organs and tissues that include several different types of cells, and even more so of entire organisms.

During rapid cooling (for example, immersing a sample in liquid nitrogen), water does not have time to diffuse out of the cells; crystals form both outside and inside cells, but due to faster cooling they turn out to be much smaller than in the first case, and do not have time to form in all cells. In this case, toxic concentrations of salts can be avoided, and the duration of their exposure is shorter, as is the duration of the harmful effects of cryoprotectors. The latter allows the use of higher concentrations.

With sufficiently rapid cooling to 0 °C and slightly lower, water does not freeze (crystallize) immediately. First, a supercooled liquid is formed. In the experiments mentioned by Smith, she was in some cases able to cool golden hamsters to -6 ° C without the formation of ice crystals. At the same time, the skin and limbs of the animals remained soft. And after warming up, the hamsters came to life without any visible harmful effects. Pregnant females (if hypothermia occurred in the first half of pregnancy) gave birth to normal cubs.

There is a technique for performing surgical operations on newborn small mammals - for example, mice. Anesthesia at this age is practically inapplicable, and therefore the cubs are simply cooled for 15–20 minutes until they lose mobility and sensitivity. There is a known case when, during such studies (the effect of removal of the vomeronasal organ on the behavior of rodents) in the laboratory of one of the Moscow institutes, several newborn Djungarian hamster cubs, due to the negligence of the experimenter, were simply forgotten lying on a cotton bedding in a chamber with a temperature of –12 ° C. After extraction - after 2-3 hours - they were completely hard, and their bodies literally “made a wooden knock”. After some time at room temperature, the cubs came to life, began to move and make sounds...

Fluids in the body usually begin to freeze at –1... –3 °C. However, as some of the water turns to ice, the concentration of solutes in the remaining liquid increases and the freezing point of that liquid continues to decrease.

The temperature of complete freezing of various biological fluids varies greatly, but in any case it turns out to be below –22...–24 °C.

The probability of the formation of a “nucleus” of an ice crystal per unit time in a supercooled liquid is proportional to the volume of this liquid and strongly depends on the temperature: at –40 ° C and at a pressure of 1 atm. crystallization of pure water occurs almost instantly, but at even lower temperatures (about -70 ° C, the rate of crystal growth slows down due to an increase in the viscosity of water. Finally, at a temperature of about -130 ° C, crystal growth stops completely. If you cool the liquid quickly enough to "overshoot" the temperature of active crystallization before crystals of a dangerous size have time to form, the viscosity increases so much that a solid glassy substance is formed. This phenomenon is called glass transition or vitrification.

If it is possible to cool cells or tissues to the glass transition temperature, they can remain in this state indefinitely, and the resulting damage will be incomparably less than during cooling with crystallization. Actually, this would be a solution to the problem of preserving biological objects in a state of deep freezing. True, when the cells thaw, in order to revive them, they will have to go through a dangerous temperature range again...

The growth rate of ice crystals in a cell can be reduced by adding impurities to the water that increase its viscosity - glycerin, sugars, etc. In addition, there are substances that block the formation of ice crystals. For example, they have such properties. special proteins produced by the organisms of a number of cold-resistant animals - Arctic and Antarctic fish, some insects, etc. The molecules of these substances have areas that are complementary to the surface of the ice crystal - “sitting” on this surface, they stop its further growth.

When cooling large (compared to a cell - 1 mm or more) objects, significant temperature gradients usually arise inside them. First, the outer layers freeze, and a so-called crystallization front is formed, moving from outside to inside. The concentration of salts and other substances dissolved in water before this front increases sharply. This leads to protein denaturation and damage to other cell macromolecules. Another problem is fabric cracking. Its cause is uneven and heterogeneous cooling, especially in a situation where the outer layers harden before the inner ones.

Back in the 60s. XX century The idea was proposed to use high pressure to control water crystallization. This idea is based on a decrease in the temperature of the water/ice phase transition with increasing pressure. At 2045 atm. The crystallization temperature of pure water is –22 °C. It is not possible to achieve a greater reduction in the freezing temperature in this way - with a further increase in pressure, it begins to rise again.

Back in 1967, an American and his colleagues experimented with freezing dog kidneys. The researchers perfused the kidneys with a 15% dimethyl sulfoxide solution (perfusion is the introduction of substances into a biological object through a system of blood vessels), and then cooled them while simultaneously increasing the pressure so that at any given moment the temperature was not below the freezing point corresponding to a given pressure. When the minimum temperature value (in this case, due to the presence of a cryoprotectant, it was about –25 °C) was reached, the pressure was reduced.

With a rapid release of pressure, a liquid supercooled to such a temperature can exist for no more than a few seconds, after which spontaneous crystallization occurs. But the crystals formed in this case are evenly distributed throughout the volume of the sample, and a crystallization front does not occur, as well as an uneven increase in the concentration of salts. In addition, the crystals that arise in this case are small in size and granular in shape and therefore cause relatively little damage to cells.

However, during the crystallization process, a significant amount of heat is released (latent heat of crystallization), as a result of which the sample is heated - ultimately to the crystallization temperature, i.e., when the pressure is reduced to atmospheric pressure - to approximately 0 ° C. After which the freezing process naturally stops. As a result, when the pressure was removed, only about 28% of the water had time to crystallize, and the rest remained liquid.

In order for all the water to crystallize, it would be necessary to cool the sample to a temperature of approximately -80 ° C before reducing the pressure - however, in this case, ice would begin to form much earlier. M. Persidski solved the problem by applying pressure cyclically. The sample, which had warmed up to 0 °C after the first release of pressure, began to be cooled again - simultaneously with a repeated increase in pressure. At its next “reset,” the next portion of the liquid had time to freeze, etc. As a result, it was possible to achieve almost complete and “harmless” crystallization of water, after which the temperature could be safely lowered to
–130 °C (and below) at normal atmospheric pressure and keep the kidney in this state for an indefinitely long time.

When thawing, the cycle was repeated in the reverse order: the kidney was heated to –28 °C, after which the pressure was increased to 2000 atm. In this case, a relatively uniform melting of ice crystals occurred. Then the sample was gradually heated with a simultaneous decrease in pressure.

Kidneys preserved in this way, according to the authors of the experiment, “showed fewer signs of tissue damage than kidneys frozen by any other method” - although they did not remain viable...

Subsequently, the high-pressure freezing technique was used in the preparation of biological samples for microscopic studies. In order to make a sufficiently thin section, the sample must first be converted into a solid state, but with conventional freezing, the cell structures are so damaged that there is practically nothing to study...

Pressure of several thousand atmospheres is successfully used for freezing products in the food industry. In this case, two goals are pursued. Firstly, after long-term (and therefore at the lowest possible temperature) storage, the taste of a frozen product should differ as little as possible from fresh. For this, it is also important that the cells are not destroyed during freezing, which can be achieved to a certain extent by freezing at a pressure of about 2 thousand atm. Another goal is the simultaneous sterilization of the product, which is achieved, on the contrary, by destroying the cells of bacteria present in it. For this, a much higher pressure is required - 6 thousand atm. and more.

The authors are not aware of new attempts to use high pressure for the reversible preservation of organs or entire organisms, and yet this path seems very promising. Of course, the question arises about the damaging effects of high pressure. It is known that with its gradual increase to approximately 500 atm. cell viability is not reduced. At 6000 atm. and more, almost all cells die, but intermediate values ​​can have different effects, depending on the type and condition of the cells, the content of water, salts and other substances in them, temperature, etc.

However, it can be expected that a gradual increase in pressure to the required 2 thousand atm. will not cause damage to the body. Indeed, in preparation for freezing, the object is first cooled to approximately 0 ° C (if it is a living creature, it stops breathing) and placed in a chamber filled with liquid. In 1961, American researcher S. Jacob subjected it to pressure of about 1000 atm for 30 minutes. the heart of a dog, just taken out of the body and continuing to contract. After the pressure was relieved, the heartbeat resumed.

It is also important that some cryoprotective substances also act as baroprotectors, i.e., they protect cells from exposure to high pressure. A “good” cryoprotector not only reduces the freezing point of the solution, but also stabilizes cell membranes, making them more elastic.

Of course, it is necessary to solve a number of problems: during experiments, to work out the optimal cooling mode, to select specific cryoprotectors, etc. For example, when going through the cycles “compression with cooling - pressure release,” cooling occurs only from the surface of the object. This leads to the fact that ice will form on the periphery, while in the center, on the contrary, existing ice may melt due to increased pressure. This can be combatted by either lowering the temperature more slowly (and allowing the object to cool more evenly) or by increasing the concentration of cryoprotective substances in the outer layers. In this case, it is not necessary to increase the pressure to maximum values. It is possible, by increasing the number of cycles, to remain within the known safe limits of 500–1000 atmospheres.

In addition, as Smith's experiments with golden hamsters showed, vitrification of only about 40% of the water (and crystallization of the rest) may be sufficient for reversible cryopreservation.

So the available data fully allow us to hope for the use of high pressures to control the crystallization of free water and cryopreservation of large biological objects, organs, and even entire organisms. Work in this direction is being carried out at the Institute of Cell Biophysics of the Russian Academy of Sciences (Laboratory of Cryopreservation of Genetic Resources under the direction) together with the Institute of Biomedical Technologies and the State Research Institute of VT named after. .

If some people geography lessons can only be seen in a nightmare, but for me it’s not like that. I enjoy reading scientific literature, I am well versed in geographical maps, and they can easily help any student. One day I realized that knowledge must be constantly developed. Now I will explain how I came to this conclusion.

I came to the sea in October to undergo wellness treatments. It was very cold that day, but I went outside for a walk near the shore. But, looking at the smart watch, I was surprised: the temperature was 0°C. Then why didn't the water in the sea freeze? Today we will find the answers together.

At what temperature does water freeze

The good thing about science is that you can find any answer. You just started thinking, but the answer is already ready. You just need to be more curious, read more books. Scientists have long announced the theory that water freezes at temperatures 0°C. Similar information is available in textbooks for schoolchildren. But this is not true. Because water solidifies, not freezes. Process turning water into ice called crystallization(this is a more accurate term).

When the temperature reaches 0°C, the water begins to change shape. Accordingly, it does not freeze completely, but only begins to freeze. It is worth considering the composition of the liquid, if there is an impurity in the water(salt, sand, dust), it will take longer to harden. There is no basis for building a crystal structure, the freezing process slows down.

Stop believing in myths about water

It is easier to remember a few statements than to read encyclopedias and conduct experiments. Therefore, even in the 21st century, people are guided by false judgments.

Top famous myths about water:

• Distilled water- the best for drinking. In fact, the cleaning process destroys everything, including beneficial minerals.

• Water- colorless substance. The water not only has transparency (may be cloudy), but also has a tint; groundwater has a yellowish or grayish tint. Sea water can be blue or dark blue.

• You can drink water in unlimited quantities. There is a formula that determines daily fluid intake that a person should drink. It all depends on weight(minimum two liters per day).

Water - Source of Life. She has power, you need to carefully use this gift of nature.

The graph (see Figure 3) also shows that the boiling point of the solution is higher than the boiling point of a pure solvent. The boiling point is the temperature at which the saturated vapor pressure is equal to the external pressure. Therefore, it is different: for pure water it is temperature T1, and for a solution it is T2 .

For dilute solutions, as the temperature decreases (see Figure 3), the pure solvent begins to crystallize first. This occurs when the vapor pressure above the solution becomes equal to the saturated vapor above the crystal (line O–B). The temperature at which crystallization begins for a solution of composition C 1 corresponds to temperature T 3, and for composition C 2 – T 4 . As the concentration of the dissolved substance increases, the freezing point decreases, which is also clearly visible on the P–T diagram (see Figure 3) .

The more concentrated the solution, the further away the vapor pressure curves over solutions are from the corresponding water curve. Therefore, the higher the concentration of the solution, the greater the difference between the boiling or freezing temperatures of water and the solution.

While studying the freezing and boiling of solutions, Raoult found that for dilute solutions of non-electrolytes, the increase in boiling point and decrease in freezing point are proportional to the concentration of the solution.

Raoult's second law:the increase in the boiling point (lower freezing point) of the solution compared to the boiling (freezing) point of the solvent is proportional to the molar concentration of the solute.

Mathematically, these temperature changes can be calculated using the formulas:

;

;

Where K E – ebulioscopic (from Latin ebullire - boil) solvent constant; To the Kyrgyz Republic– cryoscopic (from the Greek сrios - cold) constant of the solvent; – increase in boiling point; – decrease in freezing temperature; With m– molal concentration of the dissolved substance.

If you paint With m, then the formulas will take the form:

Physical meaning The ebullioscopic and cryoscopic constants are determined as follows. Their numerical values ​​show , How many degrees higher does a one-molar solution (containing 1 mole of solute in 1000 g of solvent) freeze and how many degrees below does it freeze compared to the boiling and freezing points of a pure solvent. Units of measurement are 1 deg mol -1 kg.

The ebullioscopic and cryoscopic constants do not depend on the nature of the solute, but are characteristics of the solvent. Their values ​​for some solvents are given in Table 1.

Table 1 - Cryoscopic and ebullioscopic constants of some solvents

Ebulioscopic and cryoscopic methods for determining the molecular masses of substances are based on measurements of the boiling and freezing temperatures of solutions. These two methods are widely used in chemistry, since, using various solvents, the molecular weights of various substances can be determined.

To determine the molar mass of a solute, it is convenient to use the following relationship:

Where – an increase in the boiling point or a decrease in the freezing point of the solution compared to the corresponding characteristics of a pure solvent;

K – ebullioscopic or cryoscopic constant.

The ability of solutions to freeze at a lower temperature than the solvent is used in the preparation of low-freezing solutions, which are called antifreeze. Antifreeze is used to replace water in radiators of automobile and aircraft engines in winter. The so-called main components can be used. polyhydric alcohols - ethylene glycol and glycerin:

CH 2 - CH 2 CH 2 - CH 2 - CH 2

HE HE HE HE HE

ethylene glycol glycerin

An aqueous solution of ethylene glycol (58 percent by weight), for example, freezes only at a temperature of minus 50 °C.

Osmosis

The spontaneous passage of a solvent through a semipermeable membrane separating a solution and a solvent or two solutions with different concentrations of the solute is called by osmosis . Osmosis is caused by the diffusion of solvent molecules through a semi-permeable partition, which allows only solvent molecules to pass through. Solvent molecules diffuse from a solvent into a solution or from a less concentrated solution to a more concentrated one, so the concentrated solution is diluted, and the height of its column h also increases (Figure 4).

At equilibrium, external pressure balances osmotic pressure. In this case, the rates of forward and reverse transitions of molecules through the semi-permeable partition become the same. If the external pressure applied to a more concentrated solution is higher than the osmotic p, i.e. p > p, then the rate of transition of solvent molecules from the concentrated solution will be greater, and the solvent will move into a dilute solution (or pure solvent). This process, called reverse osmosis , used to purify natural and waste waters, to obtain drinking water from sea water.

Quantitatively, osmosis is characterized by osmotic pressure, equal to the force per unit surface area, which forces solvent molecules to penetrate a semi-permeable partition. Osmotic pressure increases with increasing solute concentration and temperature. van't Hoff suggested that for osmotic pressure we can apply the equation of state of an ideal gas:

Where p– osmotic pressure, kPa; With- molar concentration of the solution, mol/l; R– universal gas constant, T– absolute temperature.

Osmosis plays a very important role in biological processes, ensuring the flow of water into cells and other structures. Solutions with the same osmotic pressure are called isotonic. If the osmotic pressure is higher than the intracellular pressure, then it is called hypertonic, if it is lower than the intracellular pressure, it is called hypotonic. For example, the average osmotic pressure of blood at 36 °C is 780 kPa. Hypertonic solutions of sugar (syrup) and salt (brine) are widely used for food preservation, as they cause the removal of water from microorganisms.

Examples of problem solving

Before solving problems, you should understand the following:

– the freezing point of the solution is lower than the freezing point of the solvent

– the boiling point of the solution is higher than the boiling point of the solvent

– value always positive and temperature change on the Celsius scale and thermodynamic Kelvin scale numerically coincide, i.e. and .

Example 1. Determination of the boiling point and freezing point of a non-electrolyte.

Determine the boiling point and freezing point of a 2% solution of naphthalene (C 10 H 8) in benzene.

Solution

Based on Raoult's second law, we can write:

We take the value of the ebullioscopic constant of benzene, as well as the boiling and freezing point of benzene from Table 1. M (C 10 H 8) = 128 g/mol. Let us remember that the percentage concentration shows the number of grams of dissolved substance in 100 g of solution, which means the mass of naphthalene is 2 g, and the mass of the solvent, i.e. benzene, is 100 – 2 = 98 g. Then, substituting the known values ​​into the equation, we get

Since pure benzene boils at 80.1 °C and the temperature rise is 0.4 °C, the boiling point of a solution of naphthalene in benzene is 80.5 °C.

The freezing point of this solution is determined in the same way:

The freezing point of benzene is 5.5 °C. The temperature decrease is 0.8 degrees, therefore, the freezing point of a 2% solution of naphthalene in benzene is 4.7 °C.

Example 2. Determination of non-electrolyte concentration based on the crystallization (boiling) temperature of solutions.

Determine the mass fraction of sucrose C 12 H 2 20 11 in water if it is known that the freezing point of this solution is minus 0.21 °C.

Solution.

From the problem data it follows that hail To determine the mass fraction of sucrose in the solution, we use the equation

into which we substitute the known values: K KR - cryoscopic constant, K KR = 1.86 deg mol -1 kg, and the molar mass of sucrose M(C 12 H 2 20 11) = 342 g/mol. Attitude

is the mass of solute per 1000 g of solvent, then

There are 38.6 g of sucrose per 1000 g of solvent, so to determine the mass fraction of the dissolved substance, you can use the formula

or make a proportion:

1038.6 g of solution contains 38.6 g of sucrose;

100 g of solution – xg sucrose.

Therefore, the mass fraction of solute is 3.71%.

Example 4. Determination of the molar mass of a non-electrolyte from the crystallization (boiling) temperature.

A non-electrolyte solution contains 2.5 g of solute in 25 g of benzene and freezes at 4.3 °C. Determine the molar mass of the solute.

Solution

Using these problem conditions and the freezing point of benzene plus 5.5°C, we determine hail The molar mass of a solute can be determined from the relation

where K KR is the cryoscopic constant of benzene, K KR = 5.12 deg mol -1 kg.

g/mol.

Control questions

1 What pressure is called saturated vapor pressure?

2 Write down the mathematical expressions for each of Raoult’s laws and explain the physical meaning of the quantities included in these expressions.

3 Equal amounts of urea CO(NH 2) 2 and sucrose C 12 H 22 O 11 were dissolved in the same amount of water under the same conditions. For which solution will the value be greater?

4 Is the decrease in the freezing point of 0.1 M aqueous solutions of glucose C 6 H 12 O 6 and urea CO (NH 2) 2 the same?

Problem 1. By how many degrees will the boiling point of an aqueous solution of urea CO(NH 2) 2 increase if 8.5 g of the substance is dissolved in 300 g of water?

Task 2. Calculate the mass fraction of methanol CH 3 OH in an aqueous solution whose freezing point is minus 2.79 °C.

Problem 3. Determine the boiling point of a solution of 1 g of naphthalene C 10 H 8 in 20 g of ether, if the boiling point of ether is 35.6 °C, K E = 2.16 °C.

Problem 4. A solution of 1.05 g of non-electrolyte in 30 g of water freezes at
– 0.7°C. Calculate the molecular weight of the nonelectrolyte.

Task 5. Calculate the amount of ethylene glycol C 2 H 4 (OH) 2 that must be added to each kilogram of water to prepare antifreeze with a freezing point of minus 15 ° C.

Task 7. To prepare antifreeze, 9 liters of glycerol C 3 H 5 (OH) 3 were taken per 30 liters of water. What is the freezing point of prepared antifreeze? The density of glycerin is 1261 kg/m3.

Dispersed systems

Chemicals can be found in pure form or in mixtures. Mixtures, in turn, can be divided into homo- and heterogeneous. Homogeneous single-phase mixtures include true solutions (see Section 1), in which the dissolved substance is presented in the form of molecules or ions, the sizes of which are comparable to the molecules of the solvent and do not exceed 1 nm. Homogeneous mixtures are thermodynamically stable.

As the particle size increases, the system becomes heterogeneous, consisting of two or more phases with a highly developed interface. And, as practice shows, another area of ​​fragmentation of matter forms a new set of properties inherent only to this form of organization of matter.

Fig.69. Curves of changes in vapor pressure of water, ice and solution depending on temperature

All pure substances are characterized by strictly defined temperatures (or points) of freezing and boiling. Thus, pure water at normal atmospheric pressure freezes at 0° and boils at 100°; benzene freezes at 5.5° and boils at 80.1°, etc. These temperatures remain unchanged until all the liquid freezes or turns into steam.

The situation is different with solutions. The presence of a solute increases the boiling point and lowers the freezing point of the solvent, and the more concentrated the solution, the more strongly. Therefore, solutions freeze at lower temperatures and boil at higher temperatures than pure solvents. It is not difficult to prove that this is a direct consequence of a decrease in the vapor pressure of solutions.

As is known, any liquid begins to boil at a temperature at which the pressure of its saturated vapor reaches the value of the external pressure. For example, at a pressure of 760 mm rt. Art. boils at 100° because at this temperature the water vapor pressure is exactly 760 mm. If you dissolve any substance in water, its vapor pressure will decrease. To bring the vapor pressure of the resulting solution to 760 mm, Obviously, you need to heat the solution above 100°. It follows that the boiling point of a solution will always be higher than the boiling point of a pure solvent.

The lower freezing point of the solution, compared to a pure solvent, is explained by the fact that the freezing point is the temperature at which the solid and liquid phases of Zhdanny can simultaneously exist (p. 218). However, for this it is necessary that the vapor pressure of the solid and liquid phases be the same, otherwise the vapor will move from one phase to another until the complete disappearance of the one above which the vapor pressure is greater. Ice and can exist together for an indefinitely long time at 0° precisely because at 0° the vapor pressure of ice (4.6 mm) equal to the vapor pressure of water. This temperature is the freezing point of pure water.

If we take some solution rather than pure water, its vapor pressure at 0° will be less than 4.6 mm; therefore, ice dipped into such a solution quickly melts. The simultaneous existence of ice and solution will be possible only at a temperature below 0°, and precisely at one at which the pressure of their vapors becomes the same. In other words, the solution will freeze at a lower temperature than the pure solvent.

All these relationships become especially clear if they are depicted graphically by drawing curves of changes in vapor pressure with temperature. In Fig. 69 line aa 1 represents the vapor pressure curve of pure water, and the line bb 1- solution vapor pressure curve. Since at any temperature the vapor pressure of the solution is less than the vapor pressure of pure water, the line bb 1 lies below the line ah. To determine from these curves the boiling point of water and solution at some pressure, for example at 760 mm, Let us draw a straight line from the corresponding point on the ordinate axis, parallel to the abscissa axis. From points a 1 and b 1 intersection of this straight line with the vapor pressure curves, we lower the perpendiculars to the abscissa axis. Temperatures T And T 1 will correspond to the boiling points of water and solution, since at these temperatures their vapor pressure is the same. We see that the boiling point of the solution lies above the boiling point of pure water.

Line ac in Fig. 69 shows the vapor pressure curve of ice. We have already said that at freezing temperature the vapor pressure of the solid and liquid phases of the solvent or solid solvent and solution shouldbe the same. This condition is met by the points A and b intersections of curves aa 1 and bb 1 with a curve ac. The freezing temperatures of water and solution are determined as projections of points A and b to the abscissa axis. In this case, as can be seen from the figure, the temperatures T And T 1 are arranged in the reverse order, i.e., the freezing temperature of the solution is less than the freezing temperature of water.

When dilute solutions freeze, a pure solvent is first released in solid form, for example, in the case of an aqueous solution, pure ice. Since it increases as ice is released, the freezing temperature does not remain constant, but gradually decreases. However, the release of ice and a decrease in the freezing temperature occur only until it reaches a certain value specific for a given substance, at which the entire solution solidifies into a solid mass. Under a microscope it can be seen that it consists of thin layers of ice and dissolved matter in solid form. This mass is called eutectic. The temperature at which its formation occurs is called eutectic temperature, and the corresponding temperature is called eutectic concentration.

Let's take, for example, a 10% solution of table salt and start cooling it. The first appearance of ice crystals is observed around -7°. As ice is released, the concentration of the remaining solution increases and the freezing point drops lower and lower. Finally, when the NaCl concentration reaches 24.42%, the entire solution solidifies into a solid white mass - the eutectic. This occurs at a temperature of -21.2°, which is therefore the eutectic temperature for a solution of table salt.

Figure 70. Freezing curve of table salt solutions

A similar picture is observed when cooling saturated solutions, i.e., solutions in which the concentration of the dissolved substance is higher than the eutectic one. The difference in the behavior of these solutions is that when they are cooled, not ice is released first, but a dissolved substance in solid form. For example, from a solution of table salt containing 26.4% NaCl, saturated at 20°C, salt is released upon cooling. As the salt is released, the concentration of the solution decreases, and when it becomes equal to 24.42%, eutectic formation occurs (at a temperature of -21.2°). Thus, when any saturated solution is cooled after a certain amount of crystals has separated, a eutectic is eventually formed.

In Fig. Figure 70 shows the freezing curve of table salt solutions of various concentrations. The figure shows that with increasing concentration the freezing point decreases. The lowest point of the curve corresponds to the eutectic temperature of -21.2° and the eutectic concentration of 24.42% NaCl. With a further increase in concentration, the freezing point (i.e., the temperature at which the release of the solid phase begins) increases again, but now it is not water that is released from the solution in solid form, but table salt.

The eutectic temperature is the lowest of all possible freezing temperatures for solutions of a given substance. It is very different for different substances. So, for example, for: potassium nitrate the eutectic temperature is only -2.9° (with an eutectic concentration of 10.9% KNO3), for table salt -21.2°, for calcium chloride -55°, for sulfuric acid -75° and etc.

The low eutectic temperature of table salt explains the melting of ice sprinkled with salt. Ice and salt cannot exist together at temperatures above -21.2°; therefore, when mixed with salt, the ice immediately begins to melt. The ability of ice to absorb a large amount of heat during melting is used to prepare cooling mixtures, discovered by Boyle in 1665 and especially carefully studied by T. E. Lovitz. In 1792, by mixing snow with calcium chloride, Lowitz achieved cooling down to -50° for the first time. It is clear that in this way it is impossible to obtain a temperature below the eutectic.

Just as when dilute solutions freeze, the released solid phase consists of a pure solvent, so when solutions of solids in liquids boil, the resulting vapors consist of a pure solvent. Therefore, as the liquid boils away, the concentration of the solution increases and the boiling point rises until the solution becomes saturated and crystallization begins. As soon as crystallization begins, the concentration of the solution stops changing and the boiling point becomes constant.

From the quantitative side, the phenomena of freezing and boiling of solutions were studied by Raoult, who experimentally established the following provisions, known as Raoult’s laws:

1. The decrease in freezing point is proportional to the amount of substance dissolved in a given weight of solvent.

So, for example, a solution containing 5 g of sugar in 100 g of water freezes at minus 0.27°, and containing 10 g - at minus 0.54°, etc.

2. Equimolecular quantities of different substances, being dissolved in the same weight quantity of a given solvent, lower its freezing point by one and the same number of degrees.

For example, when 0.1 gram of sugar (34.2 g) is dissolved in 1000 g of water, the freezing point decreases by 0.186°. The same reduction is given by 0.1 gram molecules of glucose (18 g), 0.1 gram molecules of hydrogen peroxide (3.4 G) etc.

The decrease in freezing point corresponding (by calculation) to the dissolution of 1 gram molecule of a substance in 1000 G solvent (molecular decrease), there is a constant value for a given solvent. It is called the cryoscopic constant of the solvent. The cryoscopic constants are different for different solvents. Below are some of them.

Cryoscopic constants

Water1.86°

Benzene. . . . 5.1°

Acetic acid. . . 3.9°

Naphthalene 6.9°

Exactly similar laws were established by Raoul regarding increasing boiling points. Molecular increase in boiling point, i.e. the increase caused by the dissolution of 1 gram molecule of a substance in 1000 G solvent is called the ebullioscopic constant of the solvent.

Ebulioscopic constants

Water…0.52°

Benzene…2.53°

Ether…1.82°

Chloroform…3.61°

Mathematically, Raoult's laws can be expressed by the following equation:

∆t = K x C (1)

where ∆t- lowering the freezing point or increasing the boiling point of the solvent; WITH- number of moles of solute per 1000 G solvent; TO- proportionality coefficient equal to the cryoscopic or ebullioscopic constant of the solvent, respectively (at С=1∆t = K). Since the number of moles of a substance is equal to its weight in grams (T), divided by (M), then, replacing in the previous

equation C to m: M, we get:

∆t = K(m:M) (2)

Raoult's laws apply with the same restrictions that we talked about when setting out Van't Hoff's law: concentrated solutions and electrolyte solutions deviate greatly from these laws.

The theoretical basis for Raoult's laws was given by Van't Hoff, who pointed out their connection with the law of osmotic pressure and derived equations that make it possible to calculate by lowering the freezing point or increasing the boiling point of a solution.

Very convenient methods for determining the molecular weights of dissolved substances are based on Raoult's laws. To make the determination, take a sample of the substance being tested, dissolve it in a certain amount of solvent, and determine the resulting decrease in freezing point or increase in boiling point. From these data it is easy to calculate the solute if the cryoscopic or ebullioscopic constant of the solvent is known. Conversely, knowing the soluble substance, the cryoscopic or ebullioscopic constant can be determined in the same way.

Even M.V. Lomonosov discovered in 1764 that solutions freeze at a lower temperature than pure solvents. A decrease in the freezing point of a solution is associated with a decrease in the elasticity (pressure) of the solvent vapor above the solution (a change in the concentration of cell sap in plants towards winter).

Freezing point solution is the temperature at which the solvent crystals are in equilibrium with a solution of a given composition.

Difference Δt = t 0 ° - ti ° is called temperature decrease freezing of the solution and will be greater, the higher the concentration of the solution. This dependence is expressed quantitatively by the equation:

Δt = K С m (36)

where Δt is the decrease in the freezing temperature of the solution;

Cm—molal concentration;

K is the proportionality coefficient, called cryoscopic constant solvent or a molal decrease in the freezing point of the solution.

A research method based on measuring the decrease in the freezing point of solutions is called the cryoscopic method.

Solutions freeze at lower temperatures and boil at higher temperatures than pure solvents.

For solutions of non-electrolytes, according to Raoult's law, the decrease in the freezing point of the solution is directly proportional to the molal concentration (equation 36).

The increase in the boiling point of the solution is also directly proportional to the molal concentration:

Δt kip =EC m (37)

E - ebullioscopic constant.

The osmotic pressure of solutions is calculated using the Van't Hoff formula:

Р osm =RTC m (38)

R - universal gas constant 8.314 kJ/mol deg

T - temperature, 0 K, C m - molar concentration.

Control questions

1.What is the essence of the law of distribution?

2. Derivation of the distribution law.

3. Application of the distribution law.

4. On what condition of phase equilibrium is the derivation of the distribution law based?

5.What factors influence the value of the distribution coefficient?

6.Which extraction is more effective: single or fractional?

Tasks

Job number	m g H 2 O	m g SUCHAROSE	Job number	m g H 2 O	m g SUCHAROSE
		60	.2
		55

AT WHAT TEMPERATURE WILL SOLUTIONS CONTAINING m g WATER m g SUGAR BOIL. BUILD A GRAPH OF THE DEPENDENCE OF boiling temperature on the content of dissolved substance in solution

Job number	m g H 2 O	m g SUCHAROSE	Job number	m g H 2 O	m g SUCHAROSE
		60
		55

Job number	m g H 2 O	m g GLUCOSE	Job number	m g H 2 O	m g GLUCOSE
		4,57			10,01
					12,57
					5,56
					14,40
		8,32			11,54

DETERMINE THE FREEZING TEMPERATURE OF A SOLUTION IN m g H 2 O WHICH CONTAINS m g GLUCOSE CONSTRUCT A GRAPH OF THE DEPENDENCE OF THE FREEZING temperature on the content of the dissolved substance in the solution