LESSON

LESSON №5. COLLIGATIVE PROPERTIES OF SOLUTIONS.

A solution is a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within certain limits.

Solutions can be classified on the basis of their state: solid, liquid, or gas.

The substances making up the solutions are called components. The components of a binary solution are solute and solvent. Solvent is a component which is present in excess, in other words a solvent is a substance in which dissolution takes place. Solvent doesn’t change its physical state during reaction of dissolution. Solute is a component which is present in lesser quantity. Or solute is a substance that dissolves. In a solution, the particles are of molecular size (about 1000 pm) and the different components cannot be separated by any of the physical methods such as filtration, setting ,centrifugation, etc.)

For scientific or technical applications, a qualitative account of concentration is almost never sufficient; therefore quantitative measures are needed to describe concentration. There are a number of different ways to quantitatively express concentration; the most common are listed below. They are based on mass, volume, or both. Depending on what they are based on it is not always trivial to convert one measure to the other, because knowledge of the density might be needed to do so. At times this information may not be available, particularly if the temperature varies.

Units of concentration — particularly the most popular one, molarity — require knowledge of a substance’s volume, which unlike mass is variable depending on ambient temperature and pressure. In fact (partial) molar volume can even be a function of concentration itself. This is why volumes are not necessarily completely additive when two liquids are added and mixed. Volume-based measures for concentration are therefore not to be recommended for non-dilute solutions or problems where relatively large differences in temperature are encountered (e.g. for phase diagrams). Unless otherwise stated, all the following measurements of volume are assumed to be at a standard state temperature and pressure (for example 0 degrees Celsius at 1 atmosphere or 101.325 kPa). The measurement of mass does not require such restrictions. Mass can be determined at a precision of < 0.2 mg on a routine basis with an analytical balance and more precise instruments exist. Both solids and liquids are easily quantified by weighing. The volume of a liquid is usually determined by calibrated glassware such as burettes and volumetric flasks. For very small volumes precision syringes are available. The use of graduated beakers and cylinders is not recommended as their indication of volume is mostly for decorative rather than quantitative purposes. The volume of solids, particularly of powders, is often difficult to measure, which is why mass is the more usual measure. For gases the opposite is true: the volume of a gas can be measured in a gas burette, if care is taken to control the pressure, but the mass is not easy to measure due to buoyancy effects.

The concentration of a solution may be defined as the amount of solute present

In the given quantity of the solution.

1.     Mass percentage or volume percentage

The mass percentage of a component in a given solution is the mass of the com ponent  per 100 g of the solution.

Volume-volume percentage (sometimes referred to as percent volume per volume and abbreviated as % v/v) describes the volume of the solute in mL per 100 mL of the resulting solution. This is most useful when a liquid – liquid solution is being prepared, although it is used for mixtures of gases as well. For example, a 40% v/v ethanol solution contains 40 mL ethanol per 100 mL total volume. The percentages are only additive in the case of mixtures of ideal gases.

2. Molarity

It is the number of moles of the solute dissolved per litre of the solution. It’s represented as M or  C_M

 (М) C_M = Moles of solute / Volume of solution in litres 

or

(М) C_M  = Mass of component A/ Molar mass of A *Volume of solution in litres

The unit of molarity is mol/L, 1L = 1000 ml

Molarity (in units of mol/L, molar, or M) or molar concentration denotes the number of moles of a given substance per liter of solution. A capital letter M is used to abbreviate units of mol/L. For instance:

The actual formula for molarity is:

Such a solution may be described as “0.50 molar.” It must be emphasized a 0.5 molar solution contains 0.5 moles of solute in 1.0 liter of solution. This is not equivalent to 1.0 liter of solvent. A 0.5 mol/L solution will contain either slightly more or slightly less than 1 liter of solvent because the process of dissolution causes volume of liquid to increase or decrease.

Following the SI system of units, the National Institute of Standards and Technology, the United States authority on measurement, considers the term molarity and the unit symbol M to be obsolete, and suggests instead the amount- of – substance concentration (c) with the units mol/m³ or other units used alongside the SI such as mol/L. This recommendation has not been universally implemented in academia or chemistry research yet.

Preparation of a solution of known molarity involves adding an accurately weighed amount of solute to a volumetric flask, adding some solvent to dissolve it, then adding more solvent to fill to the volume mark.

When discussing molarity of minute concentrations, such as in pharmacological research, molarity is expressed in units of millimolar (mmol/L, mM, 1 thousandth of a molar), micromolar (μmol/L, μM, 1 millionth of a molar) or nanomolar (nmol/L, nM, 1 billionth of a molar).

Although molarity is by far the most commonly used measure of concentration, particularly for dilute aqueous solutions, it does suffer from a number of disadvantages. Masses can be determined with great precision as balances are often very precise. Determining volume is ofteot as precise. In addition, due to a thermal expansion, molarity of a solution changes with temperature without adding or removing any mass. For non-dilute solutions another problem is the molar volume of a substance is itself a function of concentration so volume is not strictly additive.

Dilution calculation example:

How do you prepare 100ml of 0.40M MgSO4 from a stock solution of 2.0M MgSO4?

Answer:

There are two solutions involved in this problem. Notice that you are given two concentrations, but only one volume. Solution #1 is the one for which you have only concentration – the solution that is already sitting on the shelf. Solution #2 is the one for which you have both concentration and volume – the solution that you are going to prepare.

At least until you are comfortable with this type of problem, it may be helpful to write out what numbers go with what letters in our equation.

M1 = 2.0M MgSO4 ; V1 = unknown
M2 = 0.40M MgSO4 ; V2 = 100ml

A solution contains 5.7 grams of potassium nitrate dissolved in enough water to make 233 mL of solution. What is its molarity ?

Answer: The formula weight of KNO₃ is 101.103 g/mol.

3. Molality

It is the number of moles of the solute dissolved per 1000 g (or 1 kg) of the solvent. It’s denoted by m or C_m

(m) C_m = Moles of solute/Weight of solvent in kg

or

(m) C_m= Moles of solute * 1000/Weight of solvent in gram

The unit of Molality is m or mol/kg

 Molality is considered better for expressing the concentration as compared to molarity because the molarity changes with temperature because of expansion of the liquid with the temperature

Molality (mol/kg, molal, or m) denotes the number of moles of solute per kilogram of solvent (not solution). For instance: adding 1.0 mole of solute to 2.0 kilograms of solvent constitutes a solution with a molality of 0.50 mol/kg. Such a solution may be described as “0.50 molal”. The term molal solution is used as a shorthand for a “one molal solution”, i.e. a solution which contains one mole of the solute per 1000 grams of the solvent.

Following the SI system of units, the National Institute of Standards and Technology, the United States authority on measurement, considers the unit symbol m to be obsolete, and suggests instead the term ‘molality of substance B’ (m_B) with units mol/kg or a related unit of the SI. This recommendation has not been universally implemented in academia yet.

Note: molality is sometimes represented by the symbol (m), while molarity by the symbol (M). The two symbols are not to be confused, and should not be used as symbols for units. The SI unit for molality is mol/kg. (The unit m – not italicized – means meter.)

Like other mass-based measures, the determination of molality only requires a good scale, because masses of both solvent and solute can be obtained by weighing, and molality is independent of physical conditions like temperature and pressure, providing advantages over molarity. In a dilute aqueous solutioear room temperature and standard atmospheric pressure, molarity and molality will be very similar in value. This is because 1 kg of water roughly corresponds to a volume of 1 L at these conditions, and because the solution is dilute, the addition of the solute makes a negligible impact on the volume of the solution.

However, in all other conditions, this is usually not the case.

For example What is the molality of dentist’s amalgam? Recall that dentist’s amalgam is 70% mercury and 30 % copper, by mass.

Answer: Assume 100grams of amalgam and two significant figures, as in the previous example. By mass, mercury is the major component, so it must be the solvent. 100 grams of amalgam contains 70 grams of mercury, which is 0.070 kg of mercury. It also contains 30 grams of copper, which is 0.47 moles of copper. This gives

  Note that we have used the mass of the solvent rather than the mass of the mixture.

 4. Normality

It is the number of  gram equivalents of the solute dissolved per litre of the solution. It’s denoted by N or 

(N) Cn = Number of gram equivalents of solute/Volume of solution in litres

or

(N) Cn = Number of gram equivalents of solute *1000/Volume                                                                of solution in ml

Number of gram equivalents of solute = Mass of solute /                                                             Equivalent mass of solute

 Relationship between Normality and Molarity of Solutions

Normality = Molarity  *  Molar mass/Equivalent mass

Normality highlights the chemical nature of salts: in solution, salts dissociate into distinct reactive species (ions such as H⁺, Fe³⁺, or Cl^–). Normality accounts for any discrepancy between the concentrations of the various ionic species in a solution. For example, in a salt such as MgCl₂, there are two moles of Cl^– for every mole of Mg²⁺, so the concentration of Cl^– is said to be 2 N (read: “two normal”). Further examples are given below. It may also refer to the concentration of a solute in any solution. The normality of a solution is the number of gram equivalent weight of a solute per liter of its solution. The definition of normality depends on the exact reaction intended.

For example, hydrochloric acid (HCl) is a monoprotic acid and thus has 1 mol = 1 gram equivalent. One liter of 1 M aqueous solution of HCl acid contains 36.5 grams HCl. It is called 1 N (one normal) solution of HCl. It is given by the following formula:

In contrast, for sulfuric acid, which is diprotic acid, 2 N is usually 1 M, but may be defined as 2 M if pH < 2, where the once-deprotonated species, hydrogen sulfate, does not deprotonate.

5. Mole fraction

It is the ratio of number of moles of  one component to the total number of moles (solute and solven) present in the solution. It’s denoted by X. Let suppose that solution contains n_a  moles of solute and n_b  moles of the solvent. Then

For example: Dentist’s amalgam is 70% mercury and 30% copper, by mass. What is the mole fraction of copper in dentist’s amalgam?

Answer: For the purpose of assigning significant figures, we will assume that the percentages are accurate to at least 1%. Assume 100 grams of amalgam, which contains 70 grams of mercury and 30 grams of copper, each value having at least two significant figures. Convert each to moles and find X. Note that X does not have units.

   In measure analysis for the characteristic the composition of solution will use molar mass of an equivalent (equivalent mass)

Molar mass of an equivalent of element is the mass of the element which combines with or displaces 1.008parts by mass of hydrogen or 8 part by mass of oxygen or 35.5 parts by mass of chlorine:             

 E = f_equiv · M_B

The factor of equivalence (f_equiv) – number, which is demonstrated which part of matter (equivalent) can react with one atom of Hydrogen, or one electron in reduction reactions.

For example:

1. Concentrated hydrochloric acid is 31% hydrochloric acid and 69% water, by mass. If the density of concentrated hydrochloric acid is 1.16g/mL, what is its molarity?

Answer: Assume 100 grams of concentrated hydrochloric acid; this has 31 grams of HCl and 69 grams of water. The formula weight of hydrochloric acid is 36.46 g/mol, so 31 grams is 0.85moles. Find the volume of the solution.

  Find the molarity.

 2. An aqueous solution of sulfuric acid has a molarity of 18 mol/L. if its density of 1.84 g/mL, what is the mole fraction of water in the solution?

Answer: Assume one liter of solution, which has 18 moles of sulfuric acid. The formula weight of sulfuric acid is 98.08 g/mol, so 18 moles has a mass of 1800 grams (1756 with two significant figures). Use the density formula to find the mass of the solution

  If 1800 grams of the solution is sulfuric acid, 40 grams must be water. The formula weight of water is 18.015 g/mol, so 40 grams of water is 2 moles. The mole fraction of water is

            An aqueous solution of hydrogen peroxide, H₂O₂, is 16.9 mol/L. if it has a density of 1.196 g/mL, what is its molality?

Answer: Assume one liter of solution; this has 16.9 moles of hydrogen peroxide. The formula weight of hydrogen peroxide is 34.015 g/mol, so the liter of solution contains 575 grams of hydrogen peroxide. Use the density formula to find the mass of one liter of solution:

 3. If the mixture has 575 grams of hydrogen peroxide in 1196 grams of solution, it must have 621 grams of water (or 0.621 kg of solvent). The molality is

4.   25mL of 6M sodium hydroxide are diluted to 1.3 L. What is the resulting concentration?

Answer:

5. A student wishes to make 0.5 L of a 3.0 M sulfuric acid solution using concentrated sulfuric acid, 18 M. How should she do this?

Answer: Very Carefully! Diluting concentrated acids is dangerous. The process releases so much thermal energy that the water used in the dilution will boil, which may cause boiling hot acid burns. Ouch!

The amount of acid to use can be calculated by the equation above.

 The student needs to carefully add 80 mL of concentrated acid to roughly 400 mL of water then add enough water to make a total of 500 mL of solution. Always add the acid to the water. It takes more energy to boil a lot of water than to boil a little acid. Besides, if something has to splash out, you want it to be the water not the acid.

To prepare the molar or the normal solution of sulfuric acid with 10% solution.
              To pour into the cylinder by the volume of 100 ml the source solution of sulfuric acid. To measure the density of the solution by an aerometer and to use her value to find her mass fraction in the handbook. To calculate the volume of an acid, which need to prepare the variants: 0,05 M; 0,1 M; 0,15 M; 0,2 M; 0,25 M; 0,05 n; 0,1 n; 0,2;1n, solution. To measure the calculated volume of the solution by a measuring cylinder and to transfer into the measuring flask at 100 ml, in which to half decanted the distilled water. To rush sulfuric acid to a water by the small portions. To mix the solution in the flask. To wait for the solution will be cold to the room temperature and to bring the level of a liquid in the flask to the mark by the distilled water.

To define the density of the prepared solution using an aerometer and to calculate the mass fraction of a solute. To compare it with a data handbooks table and to calculate the error.

Colligative properties of solutions.

What are Colligative Properties?

A we have discussed, solutions have different properties than either the solutes or the solvent used to make the solution. Those properties can be divided into two main groups–colligative and non-colligative properties. Colligative properties depend only on the number of dissolved particles in solution and not on their identity. Non-colligative properties depend on the identity of the dissolved species and the solvent.

To explain the difference between the two sets of solution properties, we will compare the properties of a 1.0 M aqueous sugar solution to a 0.5 M solution of table salt (NaCl) in water. Despite the concentration of sodium chloride being half of the sucrose concentration, both solutions have precisely the same number of dissolved particles because each sodium chloride unit creates two particles upon dissolution–a sodium ion, Na⁺, and a chloride ion, Cl^–. Therefore, any difference in the properties of those two solutions is due to a non-colligative property. Both solutions have the same freezing point, boiling point, vapor pressure, and osmotic pressure because those colligative properties of a solution only depend on the number of dissolved particles. The taste of the two solutions, however, is markedly different. The sugar solution is sweet and the salt solution tastes salty. Therefore, the taste of the solution is not a colligative property. Another non-colligative property is the color of a solution. A 0.5 M solution of CuSO₄ is bright blue in contrast to the colorless salt and sugar solutions. Other non-colligative properties include viscosity, surface tension, and solubility.

Raoult’s Law and Vapor Pressure Lowering

When a nonvolatile solute is added to a liquid to form a solution, the vapor pressure above that solution decreases. To understand why that might occur, let’s analyze the vaporization process of the pure solvent then do the same for a solution. Liquid molecules at the surface of a liquid can escape to the gas phase when they have a sufficient amount of energy to break free of the liquid’s intermolecular forces. That vaporization process is reversible. Gaseous molecules coming into contact with the surface of a liquid can be trapped by intermolecular forces in the liquid. Eventually the rate of escape will equal the rate of capture to establish a constant, equilibrium vapor pressure above the pure liquid.

If we add a nonvolatile solute to that liquid, the amount of surface area available for the escaping solvent molecules is reduced because some of that area is occupied by solute particles. Therefore, the solvent molecules will have a lower probability to escape the solution than the pure solvent. That fact is reflected in the lower vapor pressure for a solution relative to the pure solvent. That statement is only true if the solvent is nonvolatile. If the solute has its own vapor pressure, then the vapor pressure of the solution may be greater than the vapor pressure of the solvent.

Note that we did not need to identify the nature of the solvent or the solute (except for its lack of volatility) to derive that the vapor pressure should be lower for a solution relative to the pure solvent. That is what makes vapor pressure lowering a colligative property–it only depends on the number of dissolved solute particles.

summarizes our discussion so far. On the surface of the pure solvent (shown on the left) there are more solvent molecules at the surface than in the right-hand solution flask. Therefore, it is more likely that solvent molecules escape into the gas phase on the left than on the right. Therefore, the solution should have a lower vapor pressure than the pure solvent.

Figure : The Vapor Pressure of a Solution is Lower than that of the Pure Solvent

The French chemist Francois Raoult discovered the law that mathematically describes the vapor pressure lowering phenomenon. Raoult’s law is given in :

Figure %: Raoult’s Law Describes the Mathematics of Vapor Pressure Lowering

Raoult’s law states that the vapor pressure of a solution, P, equals the mole fraction of the solvent, c _solvent, multiplied by the vapor pressure of the pure solvent, P^o. While that “law” is approximately obeyed by most solutions, some show deviations from the expected behavior. Deviations from Raoult’s law can either be positive or negative. A positive deviation means that there is a higher than expected vapor pressure above the solution. A negative deviation, conversely, means that we find a lower than expected vapor pressure for the solution. The reason for the deviation stems from a flaw in our consideration of the vapor pressure lowering event–we assumed that the solute did not interact with the solvent at all. That, of course, is not true most of the time. If the solute is strongly held by the solvent, then the solution will show a negative deviation from Raoult’s law because the solvent will find it more difficult to escape from solution. If the solute and solvent are not as tightly bound to each other as they are to themselves, then the solution will show a positive deviation from Raoult’s law because the solvent molecules will find it easier to escape from solution into the gas phase.

Solutions that obey Raoult’s law are called ideal solutions because they behave exactly as we would predict. Solutions that show a deviation from Raoult’s law are called non-ideal solutions because they deviate from the expected behavior. Very few solutions actually approach ideality, but Raoult’s law for the ideal solution is a good enough approximation for the non- ideal solutions that we will continue to use Raoult’s law. Raoult’s law is the starting point for most of our discussions about the rest of the colligative properties, as we shall see in the following section.

Problem:
What is the change in vapor pressure when 164 g of glycerin (C₃H₈O₃) is added to 338 mL of H₂O at 39.8 °C.
The vapor pressure of pure H₂O at 39.8 °C is 54.74 torr
The density of H₂O at 39.8 °C is 0.992 g/mL.
Solution
Raoult’s Law can be used to express the vapor pressure relationships of solutions containing both volatile and nonvolatile solvents. Raoult’s Law is expressed by
P_solution = Χ_solventP⁰_solvent where
P_solution is the vapor pressure of the solution
Χ_solvent is mole fraction of the solvent
P⁰_solvent is the vapor pressure of the pure solvent
Step 1 Determine the mole fraction of solution
molar weight_glycerin (C₃H₈O₃) = 3(12)+8(1)+3(16) g/mol
molar weight_glycerin = 36+8+48 g/mol
molar weight_glycerin = 92 g/mol
moles_glycerin = 164 g x 1 mol/92 g
moles_glycerin = 1.78 mol
molar weight_water = 2(1)+16 g/mol
molar weight_water = 18 g/mol
density_water = mass_water/volume_water
mass_water = density_water x volume_water
mass_water = 0.992 g/mL x 338 mL
mass_water = 335.296 g
moles_water = 335.296 g x 1 mol/18 g
moles_water = 18.63 mol
Χ_solution = n_water/(n_water + n_glycerin)
Χ_solution = 18.63/(18.63 + 1.78)
Χ_solution = 18.63/20.36
Χ_solution = 0.91
Step 2 – Find the vapor pressure of the solution
P_solution = Χ_solventP⁰_solvent
P_solution = 0.91 x 54.74 torr
P_solution = 49.8 torr
Step 3 – Find the change in vapor pressure
Change in pressure is P_final – P_O
Change = 49.8 torr – 54.74 torr
change = -4.94 torr

Problem:
What is the change in vapor pressure when 52.9 g of CuCl₂ is added to 800 mL of H₂O at 52.0 °C.
The vapor pressure of pure H₂O at 52.0 °C is 102.1 torr
The density of H₂O at 52.0 °C is 0.987 g/mL.
Solution
Raoult’s Law can be used to express the vapor pressure relationships of solutions containing both volitile and nonvolitile solvents. Raoult’s Law is expressed by
P_solution = Χ_solventP⁰_solvent where
P_solution is the vapor pressure of the solution
Χ_solvent is mole fraction of the solvent
P⁰_solvent is the vapor pressure of the pure solvent
Step 1 Determine the mole fraction of solution
CuCl₂ is a strong electrolyte. It will completely dissociate into ions in water by the reaction:
CuCl₂(s) → Cu²⁺(aq) + 2 Cl^–
This means we will have 3 moles of solute added for every mole of CuCl₂ added.
From the periodic table:
Cu = 63.55 g/mol
Cl = 35.45 g/mol
molar weight of CuCl₂ = 63.55 + 2(35.45) g/mol
molar weight of CuCl₂ = 63.55 + 70.9 g/mol
molar weight of CuCl₂ = 134.45 g/mol
moles of CuCl₂ = 52.9 g x 1 mol/134.45 g
moles of CuCl₂ = 0.39 mol
Total moles of solute = 3 x (0.39 mol)
Total moles of solute = 1.18 mol
molar weight_water = 2(1)+16 g/mol
molar weight_water = 18 g/mol
density_water = mass_water/volume_water
mass_water = density_water x volume_water
mass_water = 0.987 g/mL x 800 mL
mass_water = 789.6 g
moles_water = 789.6 g x 1 mol/18 g
moles_water = 43.87 mol
Χ_solution = n_water/(n_water + n_solute)
Χ_solution = 43.87/(43.87 + 1.18)
Χ_solution = 43.87/45.08
Χ_solution = 0.97
Step 2 – Find the vapor pressure of the solution
P_solution = Χ_solventP⁰_solvent
P_solution = 0.97 x 102.1 torr
P_solution = 99.0 torr
Step 3 – Find the change in vapor pressure
Change in pressure is P_final – P_O
Change = 99.0 torr – 102.1 torr
change = -3.1 torr
Answer:
The vapor pressure of the water is reduced by 3.1 torr with the addition of the CuCl₂.

Boiling Point Elevation

One consequence of Raoult’s law is that the boiling point of a solution made of a liquid solvent with a nonvolatile solute is greater than the boiling point of the pure solvent. The boiling point of a liquid or is defined as the temperature at which the vapor pressure of that liquid equals the atmospheric pressure. For a solution, the vapor pressure of the solvent is lower at any given temperature. Therefore, a higher temperature is required to boil the solution than the pure solvent. is a phase diagram for both a pure solvent and a solution of that solvent and a nonvolatile solute that explains that point graphically.

Figure: Phase Diagram for a Solvent and its Solution with a Nonvolatile Solute

As you can see in the the vapor pressure of the solution is lower than that of the pure solvent. Because both pure solvent and solutioeed to reach the same pressure to boil, the solution requires a higher temperature to boil. If we represent the difference in boiling point between the pure solvent and a solution as ΔT_b, we can calculate that change in boiling point from the :

In the we use the units molality, m, for the concentration, m, because molality is temperature independent. The term K_b is a boiling point elevation constant that depends on the particular solvent being used. The term i in the above equation is called the van’t Hoff factor and represents the number of dissociated moles of particles per mole of solute. The van’t Hoff factor is 1 for all non-electrolyte solutes and equals the total number of ions released for electrolytes. Therefore, the value of i for Na₂SO₄ is 3 because that salt releases three moles of ions per mole of the salt.

Freezing Point Depression

As you may have noticed when we looked at the , the freezing point is depressed due to the vapor pressure lowering phenomenon. The points out that fact:

Figure: Phase Diagram for a Solution and the Pure Solvent Indicating the Freezing Point Depression

In analogy to the boiling point elevation, we can calculate the amount of the freezing point depression with the :

Note that the sign of the change in freezing point is negative because the freezing point of the solution is less than that of the pure solvent. Just as we did for boiling point elevation, we use molality to measure the concentration of the solute because it is temperature independent. Do not forget about the van’t Hoff factor, i, in your freezing point calculations.

One way to rationalize the freezing point depression phenomenon without talking about Raoult’s law is to consider the freezing process. In order for a liquid to freeze it must achieve a very ordered state that results in the formation of a crystal. If there are impurities in the liquid, i.e. solutes, the liquid is inherently less ordered. Therefore, a solution is more difficult to freeze than the pure solvent so a lower temperature is required to freeze the liquid.

Osmotic Pressure

Osmosis refers to the flow of solvent molecules past a semipermeable membrane that stops the flow of solute molecules only. When a solution and the pure solvent used in making that solution are placed on either side of a semipermeable membrane, it is found that more solvent molecules flow out of the pure solvent side of the membrane than solvent flows into the pure solvent from the solution side of the membrane. That flow of solvent from the pure solvent side makes the volume of the solution rise. When the height difference between the two sides becomes large enough, the net flow through the membrane ceases due to the extra pressure exerted by the excess height of the solution chamber. Converting that height of solvent into units of pressure (by using the ) gives a measure of the osmotic pressure exerted on the solution by the pure solvent. P stands for pressure, r is the density of the solution, and h is the height of the solution.

shows a typical setup for measuring the osmotic pressure of a solution.

Figure Setup for Measuring the Osmotic Pressure of a Solution

You can understand why more molecules flow from the solvent chamber to the solution chamber in analogy to our discussion of Raoult’s law. More solvent molecules are at the membrane interface on the solvent side of the membrane than on the solution side. Therefore, it is more likely that a solvent molecule will pass from the solvent side to the solution side than vice versa. That difference in flow rate causes the solution volume to rise. As the solution rises, by the pressure depth equation, it exerts a larger pressure on the membrane’s surface. As that pressure rises, it forces more solvent molecules to flow from the solution side to the solvent side. When the flow from both sides of the membrane are equal, the solution height stops rising and remains at a height reflecting the osmotic pressure of the solution.

The equation relating the osmotic pressure of a solution to its concentration has a form quite similar to the ideal gas law:

Although the above equation may be more simple to remember, the is more useful. This form of the equation has been derived by realizing that/ V gives the concentration of the solute in units of molarity, M.

The water – main component of human organisms and also is part of medium, in which lives the people. The main water property is solved a lot of matters with formatted solutions.

The water in organisms of the person, animal, plant is by its constituent (in a yumrn’s organism about 70 -80 % of water), solvent, and also participates in exchange reactions of matters (hydrolysis, hydration, swelling (turgescence), digestion). It executes a role of a transport system in processes of a feeding, carry of enzymes, products of a metabolism, gases, antibodies. The water is supported a condition to a homeostasis in an organism of the person (acid – alkaline, osmotic, hemodinamicil, thermal equilibrium). The water is indispensable for secrets iones, maintenance of a turgor of cages.

***

The colligative properties that we will consider in this and the next unit apply to to solutions in which the solute is non-volatile; that is, it does not make a significant contribution to the overall vapor pressure of the solution. Solutions of salt or sugar in water fulfill this condition exactly. Other solutes that have very small vapor pressures, such as iodine or ethylene glycol antifreeze, can often be considered nonvolatile in comparison to the solvent at the same temperature. Solutions in which both components possess significant vapor pressures, such as alcohol in water, will be treated in another section farther on.

1  Vapor pressure of solutions: Raoult’s law

The colligative properties really depend on the escaping tendency of solvent molecules from the liquid phase. You will recall that the vapor pressure is a direct measure of escaping tendency, so we can use these terms more or less interchangeably.

The tendency of molecules to escape from a liquid phase into the gas phase depends in part on how much of an increase in entropy can be achieved in doing so. Evaporation of solvent molecules from the liquid always leads to a large increase in entropy because of the greater volume occupied by the molecules in the gaseous state. But if the liquid solvent is initially “diluted“ with solute, its entropy is already larger to start with, so the amount by which it can increase on entering the gas phase will be less. There will accordingly be less tendency for the solvent molecules to enter the gas phase, and so the vapor pressure of the solution diminishes as the concentration of solute increases and that of solvent decreases.

The number 55.5 mol L^–1 (= 1000 g L^–1 ÷ 18 g mol^–1) is a useful one to remember if you are dealing a lot with aqueous solutions; this represents the concentration of water in pure water. (Strictly speaking, this is the molal concentration of H₂O; it is only the molar concentration at temperatures around 4° C, where the density of water is closest to 1.000 g cm^–1.)

Diagram 1 (above left) represents pure water whose concentration in the liquid is 55.5 M. A tiny fraction of the H₂O molecules will escape into the vapor space, and if the top of the container is closed, the pressure of water vapor builds up until equilibrium is achieved. Once this happens, water molecules continue to pass between the liquid and vapor in both directions, but at equal rates, so the partial pressure of H₂O in the vapor remains constant at a value known as the vapor pressure of water at the particular temperature.

In the system on the right, we have replaced a fraction of the water molecules with a substance that has zero or negligible vapor pressure — a nonvolatile solute such as salt or sugar. This has the effect of diluting the water, reducing its escaping tendency and thus its vapor pressure.

What’s important to remember is that the reduction in the vapor pressure of a solution of this kind is directly proportional to the fraction of the [volatile] solute molecules in the liquid — that is, to the mole fraction of the solvent. The reduced vapor pressure is given by Raoult’s law (1886):

From the definition of mole fraction, you should understand that in a two-component solution (i.e., a solvent and a single solute), X_solvent = 1–X_solute.

Problem Example 1

Estimate the vapor pressure of a 40 percent (W/W) solution of ordinary cane sugar (C₂₂O₁₁H₂₂, 342 g mol^–1) in water. The vapor pressure of pure water at this particular temperature is 26.0 torr.

Solution: 100 g of solution contains (40 g) ÷ (342 g mol^–1) = 0.12 mol of sugar and (60 g) ÷ (18 g mol^–1) = 3.3 mol of water. The mole fraction of water in the solution is

and its vapor pressure will be 0.96 × 26.0 torr = 25.1 torr.

Since the sum of all mole fractions in a mixture must be unity, it follows that the more moles of solute, the smaller will be the mole fraction of the solvent. Also, if the solute is a salt that dissociates into ions, then the proportion of solvent molecules will be even smaller.

Problem Example 2

The vapor pressure of water at 10° C is 9.2 torr. Estimate the vapor pressure at this temperature of a solution prepared by dissolving 1 mole of CaCl₂ in 1 L of water.

Solution: Each mole of CaCl₂ dissociates into one mole of Ca²⁺ and two moles of Cl^1–, giving a total of three moles of solute particles. The mole fraction of water in the solution will be

The vapor pressure will be 0.95 × 9.2 torr = 8.7 torr.

2  Boiling point elevation

If addition of a nonvolatile solute lowers the vapor pressure of the solution, then it follows that the temperature must be raised to restore the vapor pressure to the value corresponding to the pure solvent. In particular, the temperature at which the vapor pressure is 1 atm will be higher than the normal boiling point by an amount known as the boiling point elevation.

The exact relation between the boiling point of the solution and the mole fraction of the solvent is rather complicated, but for dilute solutions the elevation of the boiling point is directly proportional to the molal concentration of the solute:

Bear in mind that the proportionality constant K_B is a property of the solvent because this is the only component that contributes to the vapor pressure in the model we are considering in this section.

Problem Example 3

Sucrose (C₂₂O₁₁H₂₂, 342 g mol^–1), like many sugars, is highly soluble in water; almost 2000 g will dissolve in 1 L of water, giving rise to what amounts to pancake syrup. Estimate the boiling point of such a sugar solution.

Solution: moles of sucrose: (2000 g) / (342 g mol^–1) = 5.8 mol

mass of water: assume 1000 g (we must know the density of the solution to find its exact value)

The molality of the solution is (5.8 mol) ÷ (1.0 kg) = 5.8 m.

Using the value of K_b from the table, the boiling point will be raised by
(0.514 K mol^–1 kg) × (5.8 mol kg^–1) = 3.0 K,
so the boiling point will be 103° C.

3  Freezing point depression

The freezing point of a substance is the temperature at which the solid and liquid forms can coexist indefinitely — that is, they are in equilibrium. Under these conditions molecules pass between the two phases at equal rates because their escaping tendencies from the two phases are identical.

Suppose that a liquid solvent and its solid (water and ice, for example) are in equilibrium ( below), and we add a non-volatile solute (such as salt, sugar, or automotive antifreeze liquid) to the water. This will have the effect of reducing the mole fraction of H₂O molecules in the liquid phase, and thus reduce the tendency of these molecules to escape from it, not only into the vapor phase (as we saw above), but also into the solid (ice) phase. This will have no effect on the rate at which H₂O molecules escape from the ice into the water phase, so the system will no longer be in equilibrium and the ice will begin to melt .

If we wish to keep the solid from melting, the escaping tendency of molecules from the solid must be reduced. This can be accomplished by reducing the temperature; this lowers the escaping tendency of molecules from both phases, but it affects those in the solid more than those in the liquid, so we eventually reach the new, lower freezing point where the two quantities are again in exact balance and both phases can coexist .

If you prefer to think in terms of vapor pressures, you can use the same argument if you bear in mind that the vapor pressures of the solid and liquid must be the same at the freezing point. Dilution of the liquid (the solvent) by the nonvolatile solute reduces the vapor pressure of the solvent according to Raoult’s law, thus reducing the temperature at which the vapor pressures of the liquid and frozen forms of the solution will be equal.

As with boiling point elevation, in dilute solutions there is a simple linear relation between the freezing point depression and the molality of the solute:

Note that K_f values are all negative!

The use of salt to de-ice roads is a common application of this principle. The solution formed when some of the salt dissolves in the moist ice reduces the freezing point of the ice. If the freezing point falls below the ambient temperature, the ice melts. In very cold weather, the ambient temperature may be below that of the salt solution, and the salt will have no effect.

The effectiveness of a de-icing salt depends on the number of particles it releases on dissociation and on its solubility in water:

Automotive radiator antifreezes are mostly based on ethylene glycol, (CH₂OH)₂. Owing to the strong hydrogen-bonding properties of this double alcohol, this substance is miscible with water in all proportions, and contributes only a very small vapor pressure of its own. Besides lowering the freezing point, antifreeze also raises the boiling point, increasing the operating range of the cooling system. The pure glycol freezes at –12.9°C and boils at 197°C, allowing water-glycol mixtures to be tailored to a wide range of conditions.

Problem Example 4

Estimate the freezing point of an antifreeze mixture is made up by combining one volume of ethylene glycol (MW = 62, density 1.11 g cm^–3) with two volumes of water.

Solution: Assume that we use 1 L of glycol and 2 L of water (the actual volumes do not matter as long as their ratios are as given.) The mass of the glycol will be 1.10 kg and that of the water will be 2.0 kg, so the total mass of the solution is 3.11 kg. We then have:

number of moles of glycol: (1110 g) ÷ (62 g mol^–1} = 17.9 mol

molality of glycol: (17.9 mol) ÷ (2.00 kg) = 8.95 mol kg^–1

freezing point depression: ΔT_F = (–1.86 K kg^–1 mol) × (8.95 mol kg^–1) = –16.6 K so the solution will freeze at about –17°C.

Any ionic species formed by dissociation will also contribute to the freezing point depression. This can serve as a useful means of determining the fraction of a solute that is dissociated.

Problem Example 5

An aqueous solution of nitrous acid (HNO₂, MW = 47) freezes at –0.198 .C. If the solution was prepared by adding 0.100 mole of the acid to 1000 g of water, what percentage of the HNO₂ is dissociated in the solution?

Solution:

The nominal molality of the solution is (.001 mol) ÷ (1.00 kg) = 0.001 mol kg^–1.

But the effective molality according to the observed ΔT_F value is given by

ΔT_F ÷ K_F = (–.198 K) ÷(–1.86 K kg mol^–1) = 0.106 mol kg^–1; this is the total number of moles of species present after the dissociation reaction
HNO₂ → H⁺ + NO^– has occurred. If we let x = [H⁺] = [NO₂^–], then by stoichiometry, [HNO₂] = 0.100 – x and .106 – x = 2x and x = .0355. The fraction of HNO₂ that is dissociated is .0355 ÷ 0.100 = .355, corresponding to 35.5% dissociation of the acid.

4  Another view of f.p. depression and b.p. elevation

A simple phase diagram can provide more insight into these phenomena. You may already be familiar with the phase map for water, shown at the right. (For more on these, see here.)

The one shown below expands on this by plotting lines for both pure water and for its “diluted” state produced by the introduction of a non-volatile solute.

The normal boiling point of the pure solvent is indicated by point where the vapor pressure curve intersects the 1-atm line — that is, where the escaping tendency of solvent molecules from the liquid is equivalent to 1 atmosphere pressure. Addition of a non-volatile solute reduces the vapor pressures to the values given by the blue line. This shifts the boiling point to the right , corresponding to the increase in temperature ΔT_b required to raise the escaping tendency of the H₂O molecules back up to 1 atm.

To understand freezing point depression, notice that the vapor pressure line intersects the curved black vapor pressure line of the solid (ice) at , which corresponds to a new triple point at which all three phases (ice, water vapor, and liquid water) are in equilibrium and thus exhibit equal escaping tendencies. This point is by definition the origin of the freezing (solid-liquid) line, which intersects the 1-atm line at a reduced freezing point ΔT_f, indicated by .

Note that the above analysis assumes that the solute is soluble only in the liquid solvent, but not in its solid form. This is generally more or less true. For example, when arctic ice forms from seawater, the salts get mostly “squeezed” out. This has the interesting effect of making the water that remains more saline, and hence more dense, causing it to sink to the bottom part of the ocean where it gets taken up by the south-flowing deep current.

… and yet another take on freezing and boiling

Skip this section if you have not yet studied chemical thermodynamics

Those readers who have some knowledge of thermodynamics will recognize that what we have been referring to as “escaping” tendency is really a manifestation of the Gibbs free energy. This schematic plot shows how the G‘s for the solid, liquid, and gas phases of a typical substance vary with the temperature.

The rule is that the phase with the most negative free energy rules

The phase that is most stable (and which therefore is the only one that exists) is always the one having the most negative free energy (indicated here by the thicker portions of the plotted lines.) The melting and boiling points correspond to the respective temperatures where the solid and liquid and liquid and vapor have identical free energies.

The relationships shown in these plots depend on the differing slopes of the lines representing the free energies of the phases as the temperature changes. These slopes are proportional to the entropy of each phase. Because gases have the highest entropies, the slope of the “gaseous solvent” line is much greater than that of the others. (This principle is explained here.) Note that this plot is not to scale.

As we saw above, adding a solute to the liquid dilutes it, making its free energy more negative, with the result that the freezing and boiling points are shifted to the left and right, respectively.

5  Colligative properties and entropy

Skip this section if you have not yet studied chemical thermodynamics

All these effects result from “dilution” of the solvent by the added solute, and because of this commonality they are referred to as colligative properties (Lat. co ligare, connected to.) The key role of the solvent concentration is obscured by the greatly-simplified expressions used to calculate the magnitude of these effects, in which only the solute concentration appears. The details of how to carry out these calculations and the many important applications of colligative properties are covered elsewhere. Our purpose here is to offer a more complete explanation of why these phenomena occur.

Basically, these all result from the effect of dilution of the solvent on its entropy, and thus in the increase in the density of energy states of the system in the solution compared to that in the pure liquid. Equilibrium between two phases (liquid-gas for boiling and solid-liquid for freezing) occurs when the energy states in each phase can be populated at equal densities. The temperatures at which this occurs are depicted by the shading.

Dilution of the solvent adds new energy states to the liquid, but does not affect the vapor phase. This raises the temperature required to make equal numbers of microstates accessible in the two phases.

Dilution of the solvent adds new energy states to the liquid, but does not affect the solid phase. This reduces the temperature required to make equal numbers of states accessible in the two phases.

Effects of pressure on the entropy: osmotic pressure

When a liquid is subjected to hydrostatic pressure— for example, by an inert, non-dissolving gas that occupies the vapor space above the surface, the vapor pressure of the liquid is raised. The pressure acts to compress the liquid very slightly, effectively narrowing the potential energy well in which the individual molecules reside and thus increasing their tendency to escape from the liquid phase. (Because liquids are not very compressible, the effect is quite small; a 100-atm applied pressure will raise the vapor pressure of water at 25°C by only about 2 torr.) In terms of the entropy, we can say that the applied pressure reduces the dimensions of the “box” within which the principal translational motions of the molecules are confined within the liquid, thus reducing the density of energy states in the liquid phase.

Applying hydrostatic pressure to a liquid increases the spacing of its microstates, so that the number of energetically accessible states in the gas, although unchanged, is relatively greater— thus increasing the tendency of molecules to escape into the vapor phase. In terms of free energy, the higher pressure raises the free energy of the liquid, but does not affect that of the gas phase.

This phenomenon can explain osmotic pressure. Osmotic pressure, students must be reminded, is not what drives osmosis, but is rather the hydrostatic pressure that must be applied to the more concentrated solution (more dilute solvent) in order to stop osmotic flow of solvent into the solution. The effect of this pressure Π is to slightly increase the spacing of solvent energy states on the high-pressure (dilute-solvent) side of the membrane to match that of the pure solvent, restoring osmotic equilibrium.

Osmotic pressure, the fourth member of the quartet of colligative properties that arise from the dilution of a solvent by non-volatile solutes.

 What you should be able to do

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Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the green-highlighted terms in the context of this topic.

• State Raoult’s law in your own words, and explain why it makes sense.

• What do we mean by the escaping tendency of a molecule from a phase? How might we be able to observe or measure it?

• Explain why boiling point elevation follows naturally from Raoult’s law.

• Explaining freezing point depression is admittedly a bit more difficult, but you should nevertheless be able to explain how the application of salt on an ice-covered road can cause the ice to melt.

Concept Map

The solution is homogeneous thermodynamic nonperishable systems, which consist with two or stable more components.

Distinguish gaseous, fluid and solid solutions. The gas solutions are mixture of gases. Air is solution of gases of azote, oxygen, carbonic oxide (IV), water vapour and inert gases. The slurries are mixture of liquids, or solutions of solid matters and gases in liquids. Solid solutions are solid phases, which one were derivated at cooling of infrequent melts.

A true solution it is a homogeneous system of a changeable structure derivated two and more components an elemental Composition and physical characteristicss of solution is identical in all volume of solution. The characteristics of solution are its structure and concentration.

 MODERN THEORY ABOUT THE NATURE OF SOLUTIONS

The medical men are especially interested for liquors. The biological liquids (blood, the lymph, urine), wich is by complex mixtures of proteins, lipids, carbohydrates, salts. Physic-chemical regularity of interplay these miscellaneous behind properties and sizes of fragments both between itself, and with water moleculas ambient them, is extremely relevant for habitability of an organism.

During development of the doctrine about solutions two theories are designed: chemical and physical.

According to the physical theory (S. Arrenyus, V. Osvald, Ye. Vant-Hoff), the process of dissolution is esteemed as an even distribution particles of solvend in all volume of solution. The solvent is by inert medium, the moleculas of solvend and solvent do not interact between themselves.

The chemical theory (D. I. Mendelaev, I.A. Kablucov, M. S. Kurnacov) regarded solution as systems, which one were derivated from parts of solvend, solvent and non-persistent chemical combinations, which one will be derivated in solution with the help of hydrogen bindings, or electrostatic attractive forces at interplay particles of a solvent and solvend.

The modern theory of solutions integrates the physical and chemical theories regarded process formation of solutions as interplay between particles of different polarity.

SOLUTIONS OF GASES IN LIQUIDS

Colligative properties are properties of solutions that depend on the number of molecules in a given volume of solvent and not on the properties (e.g. size or mass) of the molecules. Colligative properties include: lowering of vapor pressure; elevation of boiling point; depression of freezing point and osmotic pressure. Measurements of these properties for a dilute aqueous solution of a non-ionized solute such as urea or glucose can lead to accurate determinations of relative molecular masses. Alternatively, measurements for ionized solutes can lead to an estimation of the percentage of ionization taking place.

Henry’s Law: The solubility of a gas dissolved in a liquid is proportional to the partial pressure of the gas above the liquid.

This is a statement of Henry’s law, which can be written

X = KP

where X is the equilibrium mole fraction of the gas in solution (its solubility), P is its partial pressure in the gas phase, and K is a constant of proportionality, usually called the Henry’s-law constant.

The partial pressure is a part of common pressure, which one is a share of each gas in gas mixture.

Henry’s law applies only when the concentration of the solute and its par­tial pressure above the solution are both low, that is, when the gas and its solution are both essentially ideal, and when the solute does not interact

On no-bottoms, where the external pressure increases, the dissolubility of gases in a blood is augmented. At fast ascent from depth the dissolubility sharply decreases, they are excreted by the way is bubble and seal vessels – aeroembolism.

Properties of a solution which depend only on the concentration of the solute and not upon its identity are known as colligative properties. These include vapor-pressure lowering, boiling-point elevation, freezing-point depression, and osmotic pressure. Each of these properties is a consequence of a decrease in the escaping tendency of solvent molecules brought about by the presence of solute particles. Escaping tendency is the tendency shown by molecules to escape from the phase in which they exist.

 Osmosis.

Suppose а concentrated solution of copper sulphate (deep blue in colour) is placed in а beaker and water (or а dilute solution of copper sulphate) is added slowly along the walls of the beaker without much disturbing the concentrated copper sulphate solution. The two layers are more or less well defined. Now if the beaker is allowed to stand, it is observed that after а few days, the solution in the beaker becomes uniformly blue throughout. This must be obviously due to the fact that the particles of the solute (Cu⁺² and SO₄^-2 ions) move slowly into the solvent and the molecules of the solvent (water) move into the copper sulphate solution. In other words, the particles of the solute and solvent mix spontaneously into each other. 

Now suppose the experiment is performed in а slightly different manner. Suppose the beaker is divided into two compartments, by а semi-permeable membrane i.е. а membrane which allows the solvent molecules to pass through but not the solute particles. Suppose again that copper sulphate solution is placed in one compartment and water in the other. It is observed that the level on the solution side begins to rise. This must be obviously due to the fact that greater number of solvent (water) molecules from the solvent side pass into the solution side through the semi-permeable membrane than the number of solvent molecules going into the solvent from the solution through the semi-permeable membrane. Similarly, if а concentrated solution is separated from а dilute solution by а semi-permeable membrane, there is а net flow of solvent from the dilute solution to the concentrated solution through the semi-permeable membrane.

The spontaneous mixing of the particles of the solute (present in the solution) and the solvent (present above the solution) to form а homogeneous mixture is called diffusion, just as the term is used for the spontaneous mixing of gases to form homogeneous mixtures.

The net spontaneous flow of the solvent molecules from the solvent to the solution or from a less concentrated solution to а more concentrated solution through а semi-permeable membrane is called osmosis (Greek: push).

Difference between Diffusion and Osmosis. The main points of difference between diffusion and osmosis nау be summed up as given below:

Osmosis:

1.     In osmosis, а semi-permeable membrane is used.

2.     In this process, there is only flow of solvent molecules and that too through the semi-permeable membrane.

3.     It takes place from lower concentration to higher concentration.

4.     It applies to solutions only.

5.     It can be stopped or reversed by applying pressure on the solution with higher concentration.

Difference:

1.     In diffusion, no semi-permeable membrane is used.

2.     In this process, the solvent as well as the solute molecules move directly into each other.

3.     It takes place from higher concentration со lower concentration.

4.     It takes place in gases as well as solutions.

5.     It cannot be stopped or reversed.

Semi-permeable membranes. The semi-permeable membranes (as defined above) are of two types:

(i)     Natural semi-permeable membranes е.g vegetable membranes and animal membranes which are found just under the outer skin of the animals and plants. The pig’ s bladder is the most common animal membrane used.

(ii)   Artificial semi-permeable membranes. The well known examples of the artificial semi-permeable membranes are parchment paper, cellophane and certain freshly precipitated inorganic substances е.у. copper ferrocyanide, silicates, of iron, cobalt, nickel etc. The precipitated substances have to be supported on some material and this is achieved by preparing the precipitate in the walls of а porous pot.

Osmotic Pressure – The upward movement of water taking place in Fig.1 (А) can be prevented if we apply mechanical force on top of the solution in the jar. The pressure just sufficient to stop osmotic pressure exerted by the solution in the jar will be the osmotic pressure exerted by the solution present in the jar. The osmotic pressure of а solution may thus be defined as the equivalent of excess pressure which must be applied, to the solution in order to prevent the passage of the solvent into it through а semi-permeable membrane separating the two, i.e. the solution and the pure solvent. As mentioned above, due to osmosis, there is а flow of solvent from the solvent to the solution or from the less concentrated solution to the more concentrated solution through the semi-permeable membrane. This flow of the solvent does not continue indefinitely. For example, consider an inverted thistle funnel at the mouth of which is tied а semi-permeable membrane (pig’ s bladder or cellophane). Suppose the thistle funnel is filled with sugar solution and then lowered into distilled water contained in а beaker. It is observed that the level of the solution inside the stem of the thistle funnel starts rising and then after some time, it becomes constant. The rise of level in the stem of the thistle funnel is obviously due to the net flow of solvent into the solution through the semi-permeable membrane. The constancy in the level shows а state of equilibrium i.е. as many molecules of the solvent eater into the solution through the semi-permeable membrane, the same number of solvent molecules from the solution go into the solvent through the semi-permeable membrane in the same time. The pressure exerted by the column h of the solution is called osmotic pressure. Thus: Osmotic pressure may be defined as the equilibrium hydrostatic pressure of the column set up as а result of osmosis.

Expression for the osmotic pressure. Osmotic pressure (Р) of а solution is found to be directly proportional to the concentration (С) of the solution and its temperature (Т). Mathematically,

Р µС;   µТ;   Р µС·Т or P=R·C·T

where R is а constant (called solution constant) and its value is found to be same as that of the “Gas constant”. The above equation is usually written as

Р = CRT ;    p_osmotic = C_osmotic RT;

Since molarity equals the number of moles of solute (n₂) per liter of solution. V, that is. Since: C = n₂/V; 

PV= nRT  – van’t Hoff equation for dilute solutions.

Measurement of osmotic pressure

 The osmotic pressure of а solution can be measured by many methods, but only two methods will be described.

1. Pfeffer’s method – А very simple apparatus was used by Pfeffer for this purpose. А battery pot with, а semipermeable membrane deposited in its wall is cemented to, а wide glass tube which ends in а thin tube at the top and carries а manometer in the side. The manometer is closed at its upper end and is filled with Hg and N₂. The solution under investigation is introduced into the pot through this tube, The apparatus is then made airtight by sealing off the tube at the top. А portion of the pot is immersed in distilled water kept at. а constant temperature. In the course of а few days, the manometer registers the maximum pressure, which is the osmotic pressure of the solution.

2. Freezing point determination method – It ha been found that there is а decrease of 1.8б⁰С in the freezing point of а solution when its osmotic pressure is0equal to one osmole. This method is much more rapid and accurate than Pfeffer’s method. A special apparatus is used to determine the freezing point о f the solution under investigation which is then compared with freezing point of the pure. solvent.

The decrease in the freezing point of the solution is one of the colligative properties of colloidal solutions. The other colligative properties e.g. elevation of boiling point and depression of the vapor density can also be used in the determination of the osmotic pressure of а solution.

Laws of osmotic pressure – These are the same as gas laws and apply to dilute solutions which occur in the living body.

1. The osmotic pressure is directly proportional to the concentration of the solute. For example, 1 % NaCI solution will have double the osmotic pressure of 0.5 % NaCl solution. The osmotic pressure of а solution is dependent upon only the number of dissolved or dissociated particles per unit volume and is independent of chemcial nature of particles. Thus а sodium ion (at. wt. 23), а molecule of glucose (mol. wt. 180) and а molecule of serum albumin (mol. wt. about 70,000) will exert equal osmotic pressure.

2. The osmotic pressure of а solution is directly proportional to the absolute temperature. For dilute solutions, the osmotic pressure is equal to the value CRT, where С = molar concentration, R = gas constant (0.082 liter atmosphere per degree per mole) and T = absolute temperature. А molar solution of a non-electrolyte (е.g. glucose or urea) at 0⁰С (or 273⁰ absolute) will exert an osmotic pressure equal to CRT = 1 х 0.082 x 273 = 22.4 atmospheres = 22.4 х 760 mm Hg = 17024 mm Hg.

The situation however is different for electrolytes; their molecules in solution are split into more than one particles or ions. For example, NaCl ionizes into Na⁺ and Cl^–. Thus а molar solution of NaCl will have double the number of particles as compared to а molar solution of а non-electrolyte in а given volume. As already mentioned, the osmotic pressure of а solution is dependent only on the total number of particles present in it. Therefore, а molar solution of NaCl will exert double the osmotic pressure exerted by а molar solution of а non-electrolyte. In the same way, а molar solution of Na₂SO₄ will exert an osmotic pressure equal to three times the osmotic pressure of а molar solution of а non-electrolyte. In summary, in case of еlесtrоlуtеs, osmolarity = molarity х number of particles resulting from ionization of each molecule. This, however, is only а generalization; in actual practice, the osmolarity is somewhat less than that calculated as above.

The unit of osmotic pressure described above, i.e. the osmole, (expressed as osmole per liter) which is equal to 22.4 atmospheres or 17024 mm Hg is too big for use in biology. Milliosmole which is equal to 1/1,000th of an osmole, i.e. about 17 mm Hg is usually employed; Blood plasma, gastric juice, pancreatic juice, liver bile and cerebrospinal fluid exert nearly equal osmotic pressure which is about 300 (28(Ito 295) milliosmoies/liter. The serum osmolarity can be easily found out by using the formula;

Serum osmolarity in mosm/l = 2 (Serum Na⁺ + Serum К⁺, both in mmol/l) + (Serum Glucose mg/dl)/18 + (Serum Urea mg/dl)/6. As serum К⁺ level is quite 1ow in both health hand disease as compared to serum Na⁺, it can be neglected, Another term osmolality is replacing osmolarity. Whereas osmolarity is expressed as osm/l, the osmolality is expressed as osm/kg Н₂0. In  the body which has dilute fluids, both terms are virtually the same and are used interchangeably.

 VALUE OSMOS IN BIOLOGICAL PROCESSES

The blood, lymph and also all intercellular lymphs alive organisms is by aqueous solutions of moleculas and ions of many matters – organic and mineral. These solutions have definite osmotic pressure. So, the osmotic pressure of a blood of the person is value a constant and equally 7,4 105 – 7,8 105 Pa(pascal). Such high value osmotic pressure in a blood is conditioned by availability in her of a plenty of ions. High-molecular connection, mainly, proteins (albumines, the globulins), introduce 0,5 % common osmotic pressure of a blood. This part of osmotic pressure of a blood call oncotical as pressure, the value which one is equal 3,5-3,9 kPa. Oncotical pressure has large value for alive organisms. At a decrease oncotical of pressure the water goes in the party by high pressure – in a tissue, producing so called oncotical edemas of a hypodermic fat.

The osmotic pressure of a blood of the person is responded osmomolar concentration Dissoluble in plasma of matters, which one equal 0,287 – 0,0303 mol / liter.

Solutions with osmotic pressure, which is equal osmotic pressure of standard solution, is called isotonic. The solutions with osmotic by pressure are called as maximum for standard, hypertonic, and solutions with the lowest osmotic pressure hypotonic. In medical practice isotonic call solutions with osmotic pressure equal to osmotic pressure of a blood plasma. Such solution is 0, 89 % solution of sodium salt, and also 4,5 -5 % solution of a glucose. The isoosmotic solutions can be entered into an organism of the person in plenties. The hypertonic salt solutions enter in an person’s organism only in small amounts. At the introducing of a plenty hypertonic of solution the erythrocytes owing to loss of water decrease in volument and shrivel. Such phenomenon is called as a plasmolysis.

Importance of osmotic pressure of plasma proteins – The plasma proteins form а colloidal solution and are the chief colloid of the plasma. The osmotic pressure of plasma proteins which is called oncotic pressure though negligible (25 to 30 mm Hg) as compared to that of plasma crystalloids (about 5,000 mm Hg) is the main force which tends to keep the plasma water within the blood vessels. This is so because the concentration of non-colloids is almost the same in plasma and in the extracellular fluid and the osmotic pressure exerted by the non-colloids on the outside and the inside of capillaries are therefore balanced by each other. If the concentration of plasma proteins decreases markedly, water leaks into tissue spaces and the pathological condition called edema results.

Isosmotic, isotonic, hyposmotic, hypotonic, hyperosmot1c and hypertonic solutions – Isosmotic solutions are those which have the same osmotic pressure; 0.15 molar or about 0.90 % NaC1 solution in water is isosmotic with the human blood plasma. If human red blood cells are placed in this solution, they remain intact and retain their original shape and volume. 0.90% NaC1 solution is therefore isosmotic as well as isotonic with red blood cells because in this case0the amount of water entering the cells is equal to that leaving them; thus there is no net gain or loss of water by the cells.

If а solution of NaCl more concentrated than 0.90 % is used to suspend the human or other mammalian red blood cells, water will leave the cells and the cells shrink and become crenated. Such solutions are called hypertonic. On the other hand, if red blood cells are suspended in hypotonic solutions which have less than 0.90 % NaC1, then water will enter the cells making them swollen and if the solution is very dilute the cells will rupture releasing their hemoglobin in the solution. The rupture of red blood cells is called hemolysis. Frog’s blood plasma is isotonic with about 0.6% NaCl solution.

It should however be noted that osmosis and tonicity are different. А hyperosmotic solution is not necessarily а hypertonic solution. Osmolarity is а function of the number of solute particles in solution while tonicity is а function of how well а given solute causes osmosis across а cell membrane. Suppose we suspend the RBCs in two solutions of different solutes equally hyperosmolar as compared to the RBC interior, to one of which the RBC membrane is impermeable but is permeable to the other. Result will be that in the first case the outside medium of the BBCs will remain hyperosmotic and water will flow out of the RBCs. Thus this solution is both hyperosmotic as well as hypertonic. In the second case some solute particles will enter the cell interior from outside, raising its osmotic pressure and the cells will not lose water. This second type of solution is therefore hyperosmotic but not hypertonic. In the same way а solution may be hyposmotic but not hypotonic.

Cell contains а fluid (cell sap) and its wall is composed of а living cytoplasmic membrane which is semi-permeable and is responsible for the phenomenon of osmosis in living organisms. If such а cell comes in contact with water or some dilute solution, the osmotic pressure of which is less than that of cell sap present in the cell, there will be а tendency of water to enter into the cell through the cell wall. The pressure developed inside the cell due to the inflow of water into it is called turgor. On the other hand, if the cell comes in contact with а solution of higher osmotic pressure, the cell would shrink due to going out of water from the cell through the cell wall. This shrinking of the cell is called plasmolysis.

Osmotic pressure creates some critical problems for living organisms. Cells typically contain fairly high concentrations of solutes, that is, small organic molecules and ionic salts, as well as lower concentrations of macromolecules. If cells are placed in а solution that has an equal concentration of solute, there will be no net movement of water in either direction. Such solutions are called isotonic. For example, red blood cells are isotonic to а 0.9% NaCI solution. When cells are placed in а solution with а lower solute concentration (i.е., а hypotonic solution), water will move into the cells. Red blood cells, for example, will swell and rupture in а process called hemolysis when they are immersed in pure water. In hypertonic solutions, those with higher solute concentrations, cells shrivel because there is а net movement of water out of the cell. The shrinkage of red blood cells in hypertonic solution (е.g., а 3% NaCI solution) is referred to as crenation.

 Because of their relatively low cellular concentration macromolecules have little direct effect on cellular osmolarity. However, macromolecules such as the proteins contain а large number of ionizable groups. The large number of ions of opposite charge that are attracted to these groups have а substantial effect on intracellular osmolarity. Unlike most ions proteins are unable to penetrate cell membranes. (Cell membranes are not, strictly speaking, osmotic membranes, since they allow the passage of various ions, nutrients, and waste products. The term dialyzing membrane gives а more accurate description of their function.) As а result, at equilibrium the concentrations for each ionic species will not be the same on both sides of а cell’s plasma membrane. Instead, the intracellular concentrations of inorganic ions will be higher than that found outside the cell. There are several consequences of this phenomenon, called the Donnan effect:

1. а constant tendency toward cellular swelling because of water entry due to osmotic pressure,

2. the establishment of an electrical gradient called а membrane potential.

Because of the Donnan effect, cells must constantly regulate their osmolarity. Living organisms use several strategies to accomplish this goal. Many cells, for example, animal and bacterial cells, pump out certain inorganic ions such as Na+. This process, which requires а substantial proportion of cellular energy, maintains cell volume within acceptable limits. Several species, such as some protozoa and algae, periodically expel water from special contractile vacuoles. Since plant cells have rigid cell walls, plants use the Donnan effect to create an internal hydrostatic pressure called turgor pressure. This process is the driving force in cellular growth and expansion. It is also responsible for the rigidity of many plant structures.

 Fig. Osmotic Pressure and Plant Cells. (а) Isotonic solutions cause no changes in cell volume. (b) Plant cells typically exist in а hypotonic environment. Water enters these cells, and they become swollen. Cell bursting is prevented by the restraining force of rigid cell walls. (с) In а hypertonic environment the cell membrane pulls away from the cell wall because of water loss. This is the reason plants wilt when they receive insufficient water.

  a)                                    b)                                c)

Example

An aqueous solution containing 1.10 g of a protein in 100 mL of solution has an osmotic pressure of 3.93 × 10^–3 atm at 25°C. What is the molar mass of the protein?

or in SI units

Note the very high molar mass of the protein.

Vapor-pressure lowering.

The escaping tendency of a liquid is measured by its vapor pressure, which is decreased by the presence of any solute. We saw that at any given temperature the vapor pres­sure of a pure liquid depends on the fraction of its molecules that have sufficient kinetic energy to escape from the attractions of their neighbors. Shows a pure liquid solvent in equilibrium with its vapor. Shows a solution at the same temperature, also in equilibrium with its vapor. (The solute is assumed to be nonvolatile, and so the only molecules in the gas phase are those of the solvent.) In this illustration, the concentration of molecules in the gas phase can be seen to be less. In the solution, not all of the molecules at the surface are solvent molecules, and so not all are potentially capable of escaping from the liquid. Thus, the rate of evaporation from the solution is less than from the pure solvent. As a result, the equilibrium concentration of molecules in the gas phase above the solution is less than that above the pure solvent. Thus, the vapor pressure of the solution is less than that of the pure solvent.

Shows two vapor-pressure curves, one for a pure solvent and one for its solution. The vertical distance between the two curves shows the magnitude of the vapour – pressure lowering at each temperature.

Raoult’s law. The relationship between vapor-pressure lowering and concentration in an ideal solution is stated in Raoult’s law.

Raoult’s Law: The partial vapor pressure of a component in liquid solution is propor­tional to the mole fraction of that component, the constant of proportionality being the vapor pressure of the pure component.

This means that a component’s vapor pressure is equal to the product of its mole fraction times its vapor pressure when pure.

Representing the solvent by the subscript 1, Raoult’s law can be written as

P₁  =  X₁ P₁⁰

where P₁ and P₁⁰ are the vapor pressure of the solution (actually, the vapor pressure of the solvent in the solution) and that of the pure solvent, respec­tively, and X₁ is the mole fraction of the solvent in the solution. Note that as long as the solute is not volatile, P₁ is the total vapor pressure of the solution. Now, since X₁ = 1 – X₂, where the subscript 2 represents the solute, P₁  = (1- X₂)P₁⁰

or, solving for X₂,

On the right-hand side of this relationship, P₁⁰ – P₁  is the vapor-pressure

lowering brought about by the presence of the solute. is called the fractional vapor-pressure lowering, which can be seen to be equal to the mole fraction of the solute.

Boiling-point elevation.

The boiling point of a liquid is the temperature at which its vapor pressure equals atmospheric pressure. At any temperature, the pres­ence of a solute lowers the vapor pressure of a liquid, and so in order to cause a solution to boil it is necessary to raise its temperature above the boiling point of the pure solvent.

Because it is a colligative property, vapor-pressure lowering in dilute so­lutions depends on the concentration of solute particles but not on their iden­tity. Therefore, we anticipate a similar relationship between boiling-point ele­vation and solute concentration. It can be shown that in dilute solutions the boiling-point elevation is proportional to the molality of the solute particles. (As before, we assume that the solute is not volatile.) In other words, if DT_b, represents the boiling-point elevation, , then: DT_b = K_bm:

Where m is the molality of the solute and K_b, is a proportionality constant known as the molal boiling-point elevation constant.

Example:

A sample of 1.20 g of a non-volatile organic compound is dissolved in 60.0 g benzene. The BP of solution is 80.96°C. BP of pure benzene is 80.08°C. What is the molar mass of the solute.

DT = 80.96 – 80.08 = 0.88°C
    

but, we only have 60.0 g benzene, not 1000 g. so # moles solute = molality × #kg solvent

NOTE: this is only an approximate molar mass, due to the inaccuracy of the measurement (small temperature effect, hard to measure accurately).

Freezing-point depression.

The phenomenon of boiling-point elevation occurs because the presence of a solute lowers the escaping tendency of the solvent. Therefore, in order to cause a solution to boil, it is necessary to raise its temperature above the boiling point of the pure solvent. Escaping tendency means tendency for molecules to escape to any other phase, however. Consequently, in order to cause solvent molecules to freeze out of a solution, the solution must be cooled to a temperature lower than the freezing point of the pure solvent so as to compensate for the decreased escaping tendency of its molecules. The presence of a solute always lowers the freezing point of a solvent, as long as the solute is not also soluble in the solid­ified solvent. The depression of the freezing point causes the solid-liquid equi­librium line to be moved to the left in the phase diagram. Shows a composite representation of two phase diagrams: the first is for pure H₂O, and the second is for an aqueous solution of a solute that is not soluble in ice. Note that the solid-gas (sublimation) equilibrium line is unaffected. The diagram shows that the presence of the solute decreases the temperature at which ice and liquid can coexist at any pressure.

The relationship between freezing-point depression and molality in di­lute solutions is a direct proportionality and is similar to that between boiling-point elevation and molality: DT_f= K_fm

Where:  m  – molality of solute;   K_f – molal freezing-point depression constant

  – freezing-point depression.

 Example

A solution of 2.95 g of sulfur in 100 g cyclohexane had a freezing point of 4.18°C, pure cyclohexane has a fp of 6.50°C. What is the molecular formula of sulfur?

DT_f = 6.5 – 4.18 = 2.32°C
k_f = 20.2°C kg mol^-1 (look up in tables)

This is only the approximate molar mass of the sulfur since this technique is not very accurate (only 2 or three sig figs in this experiment).

atomic molar mass of sulfur is 32.06 g/mol. It takes 8 atoms of sulfur to add up to about 257 g/mol. Thus, the molecular formula for sulfur is S₈ and the true molar mass is 8 × 32.06 = 256.48 g/mol

Example

How much glycol (1,2-ethanediol), C₂H₆O₂ must be added to 1.00 L of water so the solution does not freeze above –20°C?

k_f (H₂O) = 1.86°C kg mol^–1
DT_f = k_f m

since 1.0 L has a mass of 1.0 kg we need 10.8 mol of ethylene glycol

10.8 mol × 62.1 g/mol = 670 g ethylene glycol.

Problem :

What is the vapor pressure of the pure solvent if the vapor pressure of a solution of 10 g of sucrose (C₆H₁₂O₆) in 100 g of ethanol (C₂H₆O) is 55 mmHg?

To solve this problem, we will use Raoult’s law:

Then rearrange the equation to solve for the pressure of the pure solvent, P^o. After converting the gram amounts to moles we find that the mole fraction of the solvent ethanol is 0.975. Therefore, the vapor pressure of the solvent is 56.4 mmHg

Problem :

What is the freezing point of a solution of 15.0 g of NaCl in 250 g of water? The molal freezing point constant, K_f, for water is 1.86 ^oC kg / mol.

ΔT_f = – i K_f m

For NaCl, i = 2. The concentration of the solution is 1.03 m in NaCl. Therefore, the change in the freezing point of the water is -3.8 ^oC. The freezing point of the solution is, therefore, -3.8 ^oC.

Problem :

A solution of 0.5 g of an unknowonvolatile, nonelectrolyte solute is added to 100 mL of water and then placed across a semipermeable membrane from a volume of pure water. When the system reaches equilibrium, the solution compartment is elevated 5.6 cm above the solvent compartment. Assuming that the density of the solution is 1.0 g / mL, calculate the molecular mass of the unknown.

To solve this problem, we will rearrange the formula for osmotic pressure:

Then we can calculate the pressure from the pressure depth equation, then convert the units into atmospheres.

Next, we can calculate the molarity of the solution. Finally, we will use that molarity to calculate the molar mass of the unknown from the volume of the solution and the mass of the unknown.

References:

 1.The abstract of the lecture.

2. intranet.tdmu.edu.ua/auth.php

3. Atkins P.W. Physical chemistry. – New York. – 1994. – P.299-307.

4. Cotton, F. A., Chemical Applications of Group Theory, John Wiley & Sons: New York, 1990

5.Girolami, G. S.; Rauchfuss, T. B. and Angelici, R. J., Synthesis and Technique in Inorganic Chemistry, University Science Books: Mill Valley, CA, 1999

6.John B.Russell. General chemistry. New York.1992. – P. 550-599

7. Lawrence D. Didona. Analytical chemistry. – 1992: New York. – P. 218 – 224.

8. en.wikipedia.org/wiki

Prepared by PhD Falfushynska H.