## Components of a Solution

Solutions are a type of homogeneous mixture in which one or materials (the solute) are dissolved in another material (the solvent). Solutions are most often composed of solid and/or gas solutes and a liquid solvent. When a solution is made of two or more liquids, it is not feasible to identify a solute and solvent. In this situation, liquids are said to be miscible (they dissolve in one another) or immiscible. Oil and vinegar provide an example of two liquids that are immiscible because they do not dissolve in one another.

## Concentration - Molarity

Quantitatively, concentration is important for determining the relative amount of solute dissolved per unit of solution. There are many units used for measuring the concentration of a solution. Molarity measures the number of moles of solute dissolved per liter of solution. It is important for stoichiometric calculations as it is one of the few units that quantifies moles of solute, and the only unit that does so in terms of the volume of solution. The formula used for calculating molarity is given below. The variable nsolute represents the the number of moles of solute, and Vsolution represents the volume of the solution, in liters or dm3. ## Problems - Molarity

Example #1 - Find the molarity of solution that can be made by dissolving 3.50 g of NaCl in 230.0 mL of solution. There are two important reminders to keep in mind here. The first is that the molarity formula does not gave a variable in which mass can be substituted. The mass of solute must always be converted to moles before using the molarity formula. The second is that the molarity formula requires that the volume is given in liters. For this reason, 230.0 mL becomes 0.2300 L. Example #2 - Determine the mass of potassium bromide needed to make 415.5 mL of a 0.185 M solution. ## Concentration - Molality

Though only one letter different than molarity, molality is a unique unit of concentration. It measures the moles of solute per kilogram of solvent. In an aqueous solution (i.e. one that has water as the solvent), the number of kilograms of solvent is equal to the volume of water used. This relationship is based the density of water, which is 1 g/mL. One liter (1000 mL) of water should have a mass of 1000 g, which is one kilogram. Therefore, one kilogram of water occupies a volume of one liter. This relationship works for water solutons only!

The formula used for calculating the molality of a solution is given below (note the molality is represented by the variable b): In this equation nsolute represents the number of moles of solute in the solution. The variable msolvent represents the mass of the solvent, in kilograms. Numerically, for aqueous solutions the molarity and molality are similar, but not identical in value.

## Electrolytes

An electrolyte is a compound that dissociates into ions when dissolved in water. The level of dissociation is important, and serves as the means of further classifying compounds as either strong or weak electrolytes. Strong electrolytes dissociate 100% in water, while nonelectrolytes exhibit partial dissociation. A nonelectrolyte is a compound that does not dissociate into ions in water. This does not necessarily mean the compound is insoluble in water. Sugar is a good example of a nonelectrolyte. It is quite soluble in water, but because it is organic in nature it is incapable if dissociating into ions.

## ΔGreaction Can Predict Solubility

Although a solubility table or solubility rules can be employed to determine if an ionic compound is soluble in water, ΔGreaction (or perhaps more appropriately in this scenario ΔGsolution) can be calculated for a dissociation reaction and used to determine solubility. A positive value will indicate the compound is not soluble in water, while a negative value indicates it is soluble in water. Additionally, the value for Ksp for a partially soluble salt can be found from ΔGreaction using the formula ΔGreaction = -RTlnK. Solving for K will give the solubility product constant.

## Colligative Properties

Colligative properties of a solution are those that depend on the ratio of solute particles (ions in many cases) to solvent particles, and not the identity of the solvent. Freezing point depression and boiling point elevation are examples of colligative properties. In other words, two solutions of equal volume, one made by dissolving 1 mol of NaCl and the other made by dissolving 1 mol of KBr, each lower the freezing point the same amount. This is because the concentration of ions dissolved in solution is the same; it has nothing to do with the identity of NaCl or KBr.

## Freezing Point Depression and Boiling Point Elevation

Specfically, it is the molality that is important when considering freezing point depression and boiling point elevation. The freezing and boiling point of a liquid can be changed when solutes are dissolved in it. For water, the freezing point can be lowered (depressed) by 1.86 degrees Celsius for per molal of solute. The boiling point is raised (elevated) by 0.51 degrees Celsius for per molal of solute. Additionally, solutes that dissociate in water yield a greater number of particles in solution, and therefore alter the freezing point more. For example, when 1 mol of sugar is dissolved in 1 kg of water (a solution that is 1 molal), the boiling point of the water is raised by 0.51 degrees Celsius to 100.51 degrees Celsius. However, if 1 mol of NaCl is dissolved in 1 kg of water, the boiling point will be raised by 1.02 degrees Celsius. This is twice the change that the sugar caused despite the fact that both solutions contained 1 mol of solute. The reason for the difference is because NaCl will dissociate into two moles of ions. Sugar, a nonelectrolyte, does not dissociate and therefore produces no ions. All electrolytes that dissociate in water will change the boiling point (or freezing point) by a multiple equal to the number of ions produced. In other words, a solution of CaCl2 will generate three moles of ions when dissolved in solution. One mole of CaCl2 dissolved in one kilogram of water will raise the boiling point by 1.53 degrees Celsius, or three times 0.51 degrees Celsius.