Question

Simple Answers Please

1. Explain/Discuss what a colligative property is. Provide at least 2 examples of colligative properties.

2. Provide at least on real-life application of freezing point depression.

Answer 1

1. Colligative Properties

Colligative properties are those properties of solutions that depend on the number of dissolved particles in solution, but not on the identities of the solutes. For example, the freezing point of salt water is lower than that of pure water, due to the presence of the salt dissolved in the water. To a good approximation, it does not matter whether the salt dissolved in water is sodium chloride or potassium nitrate; if the molar amounts of solute are the same and the number of ions are the same, the freezing points will be the same. For example, AlCl 3 and K 3 PO 4 would exhibit essentially the same colligative properties, since each compound dissolves to produce four ions per formula unit. The four commonly studied colligative properties are freezing point depression, boiling point elevation, vapor pressure lowering, and osmotic pressure. Since these properties yield information on the number of solute particles in solution, one can use them to obtain the molecular weight of the solute.

a)Freezing Point Depression

The presence of a solute lowers the freezing point of a solution relative to that of the pure solvent. For example, pure water freezes at 0°C (32°F); if one dissolves 10 grams (0.35 ounces) of sodium chloride (table salt) in 100 grams (3.53 ounces) of water, the freezing point goes down to −5.9°C (21.4°F). If one uses sucrose (table sugar) instead of sodium chloride, 10 grams (0.35 ounces) in 100 grams (3.53 ounces) of water gives a solution with a freezing point of −0.56°C (31°F). The reason that the salt solution has a lower freezing point than the sugar solution is that there are more particles in 10 grams (0.35 ounces) of sodium chloride than in 10 grams (0.35 ounces) of sucrose. Since sucrose, C 12 H 22 O 11 has a molecular weight of 342.3 grams (12.1 ounces) per mole and sodium chloride has a molecular weight of 58.44 grams (2.06 ounces) per mole, 1 gram (0.035 ounces) of sodium chloride has almost six times as many sodium chloride units as there are sucrose units in a gram of sucrose. In addition, each sodium chloride unit comes apart into two ions (a sodium cation and a chloride anion ) when

The freezing point depression of a solution containing a dissolved substance, such as salt dissolved in water, is a colligative property.

dissolved in water. Sucrose is a nonelectrolyte, which means that the solution contains whole C 12 H 22 O 11 molecules. In predicting the expected freezing point of a solution, one must consider not only the number of formula units present, but also the number of ions that result from each formula unit, in the case of ionic compounds. One can calculate the change in freezing point (Δ T f ) relative to the pure solvent using the equation:

Δ T f = i K f m

where K f is the freezing point depression constant for the solvent (1.86°C·kg/mol for water), m is the number of moles of solute in solution per kilogram of solvent, and i is the number of ions present per formula unit (e.g., i = 2 for NaCl). This formula is approximate, but it works well for low solute concentrations.

Because the presence of a solute lowers the freezing point, many communities put salt on their roads after a snowfall, to keep the melted snow from refreezing. Also, the antifreeze used in automobile heating and cooling systems is a solution of water and ethylene glycol (or propylene glycol); this solution has a lower freezing point than either pure water or pure ethylene glycol.

b) Boiling Point Elevation

The boiling point of a solution is higher than that of the pure solvent. Accordingly, the use of a solution, rather than a pure liquid, in antifreeze serves to keep the mixture from boiling in a hot automobile engine. As with freezing point depression, the effect depends on the number of solute particles present in a given amount of solvent, but not the identity of those particles. If 10 grams (0.35 ounces) of sodium chloride are dissolved in 100 grams (3.5 ounces) of water, the boiling point of the solution is 101.7°C (215.1°F; which is 1.7°C (3.1°F) higher than the boiling point of pure water). The formula used to calculate the change in boiling point (Δ T b ) relative to the pure solvent is similar to that used for freezing point depression:

Δ T b = i K b m ,

where K b is the boiling point elevation constant for the solvent (0.52°C·kg/mol for water), and m and i have the same meanings as in the freezing point depression formula. Note that Δ T b represents an increase in the boiling point, whereas Δ T f represents a decrease in the freezing point. As with the freezing point depression formula, this one is most accurate at low solute concentrations.

2. Chemistry of Ice Cream making is the example of Lowering of freezing point.