LESSON 11. Glossary: Solutions. Boiling-point elevation

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LESSON 11 Glossary: Solutions Boiling-point elevation Colligative properties Freezing-point depression Molality Molarity (M) Mole (mol) Mole fraction Saturated solution a colligative property of a solution equal to the boiling point of the solution minus the boiling point of the pure solvent properties that depend on the concentration of solute molecules or ions in a solution but not on the chemical identity of the solute a colligative property of a solution equal to the freezing point of the pure solvent minus the freezing point of the solution the moles of solute per kilogram of solvent the moles of solute dissolved in one liter (cubic decimeter) of solution the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 g of carbon- 12. The amount of substance containing Avogadro's number of molecules or formula units the moles of a component substance divided by the total moles of solution a solution that is in equilibrium with respect to a given dissolved substance

Solute Solution Solvent Supersaturated solution Unsaturated solution in the case of a solution of a gas or solid dissolved in a liquid, the gas or solid; in other cases, the component in smaller amount a mixture that is uniform in its properties throughout given samples in a solution of a gas or solid in a liquid, the liquid; in other cases, the component in greater amount a solution that contains more dissolved substance than does a saturated solution; the solution is not in equilibrium with the pure substance a solution that is not in equilibrium with respect to a given dissolved substance and in which more of the substance can dissolve Types of Solutions Helpful Link: Visit this link to view a video of solutions. http://www.backpack.tv/video/chemistry/washingtonville-high-school/markrosengarten/chemistry-tutorial-101a-solutions-and You may be asking yourself why solutions are necessary. Well, they serve a lot of practical purposes. For instance, most chemical reactions are run in solution. Solutions also usually have a lower melting and freezing point than just a substance by itself. Let's say that we have a handful of salt crystals, NaCl. When this salt dissolves in water, there is a uniform dispersion of ions in this water. What we have just

formed is called a solution. A solution is defined as a homogeneous mixture of two or more substances, consisting of ions or molecules. A solution can be a solid, liquid, or gas. A solution results from the combination of a solute and a solvent. A solute is, in the case of a gas or solid dissolved in liquid, the gas or solid, but in other cases it is the part that has the smaller amount. A solvent is, in the case of a gas or solid dissolved in liquid, the liquid, but in other cases it is the part that has the larger amount. In the example concerning the salt, the solvent would be water and the salt would be the solute. So that we can understand the concept of solubility, let's consider the process of dissolving sodium chloride in water. We know that NaCl is an ionic compound, and that it dissolves in water as Na + and Cl - ions. When we mix 50 grams of salt in 100 ml of water we find something cool going on. The sodium and chlorine ions leave the surface of the crystal and enter into the solution. These ions then move about at random in the solution and occasionally bump into another crystal and stick, returning to their crystalline state. As time passes, more sodium chloride continues to dissolve into the solution, which means that the ion concentration increases. This means that eventually more ions are going to bump into each other and return to their crystalline state. Over time, a dynamic equilibrium will be reached in which the rate at which ions leave the crystals is equivalent to the rate at which ions collide to form a crystal. At this equilibrium, no more salt appears to dissolve. We have reached what we call a saturated solution. A saturated solution is a solution that is in equilibrium with respect to a given dissolved substance. An unsaturated solution is a solution that is not in equilibrium with respect to given dissolved substance, leaving room for more substance to dissolve. The solubility of a substance is the amount that dissolves in a given quantity of water at a given temperature to give a saturated solution. Sometimes, through special circumstances we can obtain a supersaturated solution. A supersaturated solution is a solution that contains more dissolved solute than is normally possible at that temperature. Molarity Helpful link: Visit this link to view a video on molarity and molality.

http://www.backpack.tv/video/chemistry/bozeman-science/paul-andersen/solutions-andmolarity http://www.backpack.tv/video/chemistry/washingtonville-high-school/markrosengarten/chemistry-tutorial-103d-molality Molar Concentration or Molarity is defined as the moles of solute dissolved in one liter of solution. moles of solute Molarity = liter of solution Example:.0678 g of NaCl is placed in a 25.0 ml flask full of water. When the NaCl dissolves, what is the molarity of the solution?.0464 M NaCl Molality The molality of a solution is the moles of solute per kilogram of solvent. moles of solute molality = kilogram of solvent Example:.20 mol of ethylene glycol / 2.0 Kg of solvent =.10 m ethylene glycol Mole Fraction Mole fraction is the number of moles of a substance divided by the total number of moles in the solution.

moles of substance A Mole Fraction = Total Moles of solution Example: If there is 4.0 moles of HCl in 12 moles of solution then the mole fraction of HCl would be 1/3. When all mole fractions are added together the total is always 1. K f and K b K f and K b are the freezing point depression constant and boiling point elevation constant. When a solute is added to a solvent, the boiling point of the solution is always greater than the boiling point o the pure solvent. Adding a solute also lowers the freezing point. Solvent Formula Melting Point ( o C) Boiling Point( o C) K f ( o C/m) K b ( o C/m) Benzene C 6 H 6 5.455 80.2 5.065 2.61 Ethanol C 2 H 5 OH -- 78.3 -- 1.07 Water H 2 0 0.000 100.000 1.858 0.521 To determine the amount of change in boiling point you will need this equation: T b = K b * m T b change in boiling point K b boiling point elevation constant m molality of solution Then T b is added to the normal boiling point of the pure solvent. Note that the identity of the solute is not important, just its concentration (expressed in molality). Therefore, boiling point elevation is a colligative property.

To determine the freezing point of a solution, you need to calculate the decrease in freezing point caused by the addition of a solute to the solvent. Use the equation: T f = K f * m T f change in freezing point K f freezing point depression constant m molality of solution Then the T f is subtracted from the normal freezing point of the pure solvent. Freezing point depression is also a colligative property. Example: Calculate the boiling point and freezing point of a solution of.30 g of glycerol (C 3 H 8 O 3 ) in 20.0 g of water. moles glycerol = (.30 g) (1 mole / 92 g) =.0033 moles molality of solution =.0033 moles /.020 kg =.16 m T b = (.521 o C/m)(.16 m) =.083 o C Boiling point = 100.00 +.083 = 100.083 o C T f = (1.858 o C/m)(.16m) =.30 o C Freezing point = 0.00 o C -.30 o C = -.30 o C In an ionic solution, the total concentration of ions is important. Therefore, another factor (i) is included in the equations. T b = K b * m * i T f = K f * m * i

"i" is the number of ions from each formula unit. In the previous example, if NaCl had been the solute your change in boiling point and freezing point would have needed to be multiplied by 2. (i=2 because NaCl consists of a Na + ion and a Cl - ion Lesson 11 Review MOLARITY (M) Molarity = moles of solute Liter of solution Solve the problems below. 1. What is the molarity of a solution in which 58 g of NaCI are dissolved in 1.0 L of solution? 2. What is the molarity of a solution in which 10.0 g of AgN0 3 is dissolved in 500. ml of solution? 3. How many grams of KN0 3 should be used to prepare 2.00 L of a 0.500 M solution? 4. To what volume should 5.0 g of KCI be diluted in order to prepare a 0.25 M solution? 5. How many grams of CuS0 4 e 5H 2 0 are needed to prepare 100. ml of a 0.10 M solution? MOLALITY (m) Molality = moles of solute Kg of solvent 6. What is the molality of a solution in which 3.0 moles of NaCI is dissolved in 1.5 Kg of water? 7. What is the molality of a solution in which 25 g of NaCI is dissolved in 2.0 Kg of water? 8. What is the molality of a solution in which 15 g of 1 2 is dissolved in 500g of alcohol? 9. How many grams of 1 2 should be added to 750 g of CCI 4 to prepare a 0.020m solution?

10. How much water should be added to 5.00 g of KCI to prepare a 0.500 m solution? Submit your answers in the text box.