23 ppm 2.3 ppm 0.23 ppm 230 ppm

1. Answer or If density of solution = 1g/ ml • 23 ppm • 2.3 ppm • 0.23 ppm • 230 ppm

2. ppm= 2.2 An Exercise A 2.5 g sample of ground water was found to contain 5.4 mg of Zn2+ ions. What is the concentration of Zn2+ ions in a) ppb and b) ppm Answer

3. If 3.6 mg of Na+ is detected in a 200. g sample of water from Lake Erie, what is its concentration in ppm? 7.2 ppm 1.8 ppm 18 ppm 72 ppm

4. Answer 7.2 ppm 1.8 ppm 18 ppm 72 ppm mass of component in solution =  6 ppm of component 10 total mass of solution 3.6 mg 0.0036 g =  = 6 10 18 ppm 200. g 200. g

5. moles of A total moles in solution XA = Mole Fraction (X), Molarity, and Molality Concentrations are often expressed in terms of number of moles of one or more component of the solution, the three most commonly used units are mole fraction, molarity, and molality • In some applications, one needs the mole fraction of solvent, not solute—make sure you find the quantity you need! Mole Fraction (X) XA is the mole fraction of a component “A”

6. A Practice Exercise Consider a solution containing 36.5 g HCl and 144 g of water, what is the mole fraction of HCl in solution? What is the mole fraction of water in solution?

7. Check Answer

8. mol of solute L of solution M = Molarity (M) • You will recall this concentration measure from Chapter 4. • Because volume is temperature dependent, molarity can change with temperature.

9. mol of solute kg of solvent m = Molality (m) Because both moles and mass do not change with temperature, molality (unlike molarity) is not temperature dependent. Therefore molality is the unit of choice for measurement which are carried out over a wide range of temperature

10. Changing Molarity to Molality If we know the density of the solution, we can calculate the molality from the molarity, and vice versa.

11. This question cannot be answered without additional concentration information. • Yes • No

12. An Exercise:Calculation of Molality A solution is made by dissolving 4.35 g glucose (C6H12O6) in 25.0 mL of water at 25°C. Calculate the molality of glucose in the solution. Analyze: We are asked to calculate a molality. To do this, we must determine the number of moles of solute (glucose) and the number of kilograms of solvent (water). • Plan: • Use the molar mass of C6H12O6 to convert grams to moles • Use the density of water to convert millliters to kilograms • The molality equals the number of moles of solute divided by the number of kilograms of solvent Answer Because water has a density of 1.00 g/mL, the mass of the solvent is

13. PRACTICE EXERCISE What is the molality of a solution made by dissolving 36.5 g of naphthalene (C10H8) in 425 g of toluene (C7H8)? Answer: 0.670 m

14. Answer a) b) An Exercise:Calculation of Mole Fraction and Molality An aqueous solution of hydrochloric acid contains 36% HCl by mass. (a) Calculate the mole fraction of HCl in the solution. (b) Calculate the molality of HCl in the solution. A key point: consider the total mass as 100 g, then the mass of water is 64 g

15. Answer The number of moles of solute is The density of the solution is used to convert the mass of the solution to its volume: Molarity is moles of solute per liter of solution: An Exercise:Calculation of Molarity A solution contains 5.0 g of toluene (C7H8) and 225 g of benzene and has a density of 0.876 g/mL. Calculate the molarity of the solution. What is the molality of the solution? 0.24 m

16. Colligative Properties of Solutions • Colligative properties are physical properties of solutions that depend only on the number of solute particles present, not on the identity of the solute particles. • Among colligative properties are: • Vapor pressure lowering • Boiling point elevation • Melting point depression • Osmotic pressure

17. Lowering the Vapor Pressure Because of solute-solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase. Vapor pressure lowering upon addition of a nonvolatile solute to a volatile solvent

18. Raoult’s Law • Where, • The subscript “A” represents a given volatile solvent in solution • PA is the partial pressure exerted by a solvent vapor above the solution • XA is the mole fraction of the solvent A in solution • PA is the normal vapor pressure of the pure solvent A at that temperature Raoult’s law states that the vapor pressure of a volatile solvent is reduced upon addition of a nonvolatile solute. Reduction in the vapor pressure is independent of the identity of the solute (whether ionic or molecular) but it depends only on the concentration of the solute in solution.

19. Answer Example Consider a solution that is made up of water an glucose (C6H12O6) at 20 oC, where What is the partial pressure of water above the solution? The pressure of water vapor above the solution is reduced by 3.5 torr (17.5 torr - 14.0 torr) as compared to pure water

20. Assuming that the gas mixture obeys the ideal law Recall from the previous semester, Ideal gases An Ideal solution that contains more than one volatile substance Solutions sometimes have two or more volatile components, e.g. gasoline is a complex solution containing several volatile substances Generally, consider an ideal solution that contains two volatile liquids A and B, according to Raoult’s law

21. Example Vapor is more rich in benzene? Consider a mixture of benzene (C6H6) and toluene (C7H8) containing 1.0 mol and 2.0 mol of benzene and toluene, respectively. At 20 oC the vapor pressures of pure substances are: What is the partial pressure of each component above the solution? What is the total vapor pressure above the solution? What is the mole fraction of each component in the vapor above the solution?