SAT Chemistry Liquids, Solids, and Phase Changes - Water Solutions

SAT Chemistry Liquids, Solids, and Phase Changes - Water Solutions

To make molecules or ions of another substance go into solution, water molecules must overcome the forces that hold these molecules or ions together. The mechanism of the actual process is complex. To make sugar molecules go into solution, the water molecules cluster around the sugar molecules, pull them off, and disperse, forming the solution.
For an ionic crystal such as salt, the water molecules orient themselves around the ions (which are charged particles) and again must overcome the forces holding the ions together. Since the water molecule is polar, this orientation around the ion is an attraction of the polar ends of the water molecule. For example:
Once surrounded, the ion is insulated to an extent from other ions in solution because of the dipole property of water. The water molecules that surround the ion differ in number for various ions, and the whole group is called a hydrated ion.
In general, as stated in the preceding section, polar substances and ions dissolve in polar solvents and nonpolar substances such as fats dissolve in nonpolar solvents such as gasoline. The process of going into solution is exothermic if energy is released in the process, and endothermic if energy from the water is used up to a greater extent than energy is released in freeing the particle.
When two liquids are mixed and they dissolve in each other, they are said to be completely miscible. If they separate and do not mix, they are said to be immiscible.
Two molten metals may be mixed and allowed to cool. This gives a “solid solution” called an alloy.

Figure 32 shows the general sizes of the particles found in a water mixture.
The basic difference between a colloid and a suspension is the diameter of the particles
dispersed. All the boundaries marked in Figure 32 indicate only the general ranges in which r the distinctions between solutions, colloids, and suspensions are usually made.
The characteristics of water mixtures are as follows:

When a bright light is directed at right angles to the stage of an ultramicroscope, the indi¬vidual reflections of colloidal particles can be observed to be following a random zigzag path. This is explained as follows: The molecules in the dispersing medium are in motion and continuously bumping into the colloidal particles, causing them to change direction in a random fashion. This motion is called Brownian movement after the Scottish botanist Robert Brown, who first observed it.

There are general terms and very specific terms used to express the concentration of a solu¬tion. The general terms and their definitions are:


It is interesting to note that the words saturated and concentrated are NOT synonymous. In fact, it is possible to have a saturated dilute solution when the solute is only slightly soluble and a small amount of it makes the solution saturated, but the concentration of the solution is still dilute.

Specific Terms of Concentration
The more specific terms used to describe concentration are mathematically calculated.

1. PERCENTAGE CONCENTRATION is based on the percent of solute in the solution by mass. The general formula is:
Example __________________________________

How many grams of NaCl are needed to prepare 200. grams of a 10% salt solution?
10% of 200. grams = 20.0 grams of salt
You could also solve the problem using the above formula and solving for the unknown quantity.
The next two expressions depend on the fact that, if the formula mass of a substance is expressed in grams, it is called a gram-formula mass (gfin), molar mass, or 1 mole. Gram- molecular mass can be used in place of gram-formula mass when the substance is really of molecular composition, and not ionic like NaCl or NaOH. The definitions and examples are:

2. MOLARITY (abbreviated M) is defined as the number of moles of a substance dissolved in 1 liter of solution.
A 1 molar H2SO4 solution has 98. grams of H2SO4 (its molar mass) in 1 liter of the solution.
This can be expressed as a formula:
If the molarity (M) and the volume of a solution are known, the mass of the solute can be determined. First, use the above equation and solve for the number of moles of solute. Then, multiply this number by the molar mass.

Example __________________________________

How many grams of NaOH are dissolved in 200. milliliters of solution if its concentration is 1.50 M?
3. MOLALITY (abbre viated m) is defined as the number of moles of the solute dissolved in 1,000 grams of solvent.

Example __________________________________

A 1 molal solution of H2SO4 has 98 grams of H2SO4 dissolved in 1,000. grams of water. This, you will notice, gives a total volume greater than 1 liter, whereas the molar solution had 98 grams in 1 liter of solution.
Example __________________________________

Suppose that 0.25 mole of sugar is dissolved in 500. grams of water. What is the molality of this solution?
4. MOLE FRACTION is another way of indicating the concentration of a component in a solution. It is simply the number of moles of that component divided by the total moles of all the components. The mole fraction of component i is written as Xi For a solution consisting of nA moles of component A, nB moles of component B, nc moles of compo­nent C, and so on, then the mole fraction of component A is given by:
As an example, if a mixture is obtained by dissolving 10 moles of NaCl in 90 moles of water, the mole fraction of NaCl in that mixture is 10 (moles of NaCl) divided by (10 + 90) or 100 moles, giving an answer of 0.1, the mole fraction of NaCl.

In dilution problems, the expression of molarity gives the quantity of solute per volume of solution. The amount of solute dissolved in a given volume of solution is equal to the product of the concentration times the volume. Hence, 0.5 liter of 2 M solution contains
Notice that volume units must be identical.
If you dilute a solution with water, the amount or number of moles of solute present remains the same, but the concentration changes. You can use the expression:
This expression is useful in solving problems involving dilution.

Example __________________________________

If you wish to make 1 liter of solution that is 6 M into 3 M solution, how much water must be added?
M1V1 = M2V2
6M × 1L = 3M × xL
Solving this expression:
x L = 2 L. This is the total volume of the solution after dilution and means that 1 liter of water had to be added to the original volume of 1 liter to get a total of 2 liters for the dilute solution volume.
An important use of the molarity concept is in the solution of titration problems, which are covered in Chapter 11, along with pH expressions of concentration for acids.

Colligative properties are properties that depend primarily on the concentration of particles and not the type of particle. There is usually a direct relationship between the concentration of particles and the effect recorded.
The vapor pressure of an aqueous solution is always lowered by the addition of more solute. From the molecular standpoint, it is easy to see that there are fewer molecules of water per unit volume in the liquid, and therefore fewer molecules of water in the vapor phase are required to maintain equilibrium. The concentration in the vapor drops and so does the pressure that molecules exert. This is shown graphically below.
Notice that the effects of this change in vapor pressure are registered in the freezing point and the boiling point. The freezing point is lowered, and the boiling point is raised, in direct proportion to the number of particles of solute present. For water solutions, the concentra¬tion expression that expresses this relationship is molality (m), that is, the number of moles of solute per kilogram of solvent. For molecules that do not dissociate, it has been found that aim solution freezes at -1.86°C (271.14 K) and boils at 100.51°C (373.51 K). A 2 m solution would then freeze at twice this lowering, or -3.72°C (269.28 K), and boil at twice the 1 molal increase of 0.51°C, or 101.02°C (374.02 K).

Example 1 __________________________________

A 1.50-gram sample of urea is dissolved in 105.0 grams of water and produces a solution that boils at 100.12°C. From these data, what is the molecular mass of urea?
Because this property is related to the molality, then you must find the number of grams in 1,000 g of water.
Example 2 __________________________________

Suppose that there are two water solutions, one of glucose (molar mass = 180.), the other of sucrose (molar mass = 342). Each contains 50.0 grams of solute in 1,000. grams (1 kg) of water. Which has the higher boiling point? The lower freezing point?
The molality of each of these nonionizing substances is found by dividing the number of grams of solute by the molecular mass.
Therefore, their respective molalities are 0.278 m and 0.146 m. Since the freezing point and boiling point are colligative properties, the effect depends only on the concentration. Because the glucose has a higher concentration, it will have a higher boiling point and a lower freezing point. The respective boiling points would be:
Although the previous discussion referenced the quantitative effects on the boiling and freezing point of water solutions, the SAT Subject Test in Chemistry will not ask questions of that type. The test will ask questions only about the relative degree of the effect and that recognize that the effect occurs. As stated, the relative degree of the effect depends only on the number of particles dissolved. However, the number of those particles also depends on the nature of the solute.
Using a solute that is an ionic solid and that completely ionizes in an aqueous solution introduces a greater number of particles than when a nonionizing molecular solute is dis¬solved. Notice in the preceding chart that a 1 molal solution of NaCl yields a solution with 2 moles of particles because:


Thus', a 1 molal solution of NaCl has 2 moles of ion particles in 1,000 grams of solvent. In contrast, a 1 molal solution of a sugar solution has only 1 mole of molecular particles in that same mass of solvent because the sugar does not dissociate into ions. Although the colliga- tive property of lowering the freezing point and raising the boiling point depends primarily on the concentration of particles and not the type of particles, the number of particles influ¬encing the property does vary with the type of solute dissolved.

Example  __________________________________

Rank the following aqueous solutions based on the range of temperature in which the solu¬tion is a liquid. Rank the solutions from largest to smallest range.

0.50 m KI       0.50 m MgCl2       0.75 m C2H5OH      1.00 m CaCl2

The solution that has the greatest range of temperature over which the solution is a liquid is the one that has the lowest freezing point and the highest boiling point. All but the ethyl alcohol (C2H5OH) solution contain ionic solutes that would dissociate into the number of ions as dictated by their formulas. That means that the total number of moles of solute particles per 1,000.0 grams of water in each solution would be 1.00, 1.50, 0.75, and 3.00 respectively. Therefore, the solutions should be ranked as shown below:
#1. (greatest range) 1.00 m CaCl2
#2. 0.50 m MgCl2
#3. 0.50 m KI
#4. (smallest range) 0.75 m C2H5OH

The example above explains the use of salt on icy roads in the winter and the increased effectiveness of calcium chloride per mole of solute. The use of glycols in antifreeze solutions in automobile radiators is also based on this same concept.

Many substances form a repeated pattern structure as they come out of solution. The struc¬ture is bounded by plane surfaces that make definite angles with each other to produce a geometric form called a crystal. The smallest portion of the crystal lattice that is repeated throughout the crystal is called the unit cell. Samples of unit cells are shown in Figure 33. The crystal structure can also be classified by its internal axis, as shown in Figure 34.
A substance that holds a definite proportion of water in its crystal structure is called a hydrate. The formulas of hydrates show this water in the following manner: CuSO4 • 5H2O; CaSO4 • 2H2O; and Na2CO3 • 10H2O. (The • is read as “with.”) When these crystals are heated gently, the water of hydration can be forced out of the crystal and the structure collapses into an anhydrous (without water) powder. The dehydration of hydrated CuSO4 serves as a good example since the hydrated crystals are deep blue because of the copper ions present with water molecules. When this water is removed, the structure crumbles into the anhydrous white powder. Some hydrated crystals, such as magnesium sulfate (Epsom salt), lose the water of hydration on exposure to air at ordinary temperatures. They are said to be efflores­cent. Other hydrates, such as magnesium chloride and calcium chloride, absorb water from
the air and become wet. They are said to be deliquescent or hydroscopic. This property explains why calcium chloride is often used as a drying agent in laboratory experiments.


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