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Percent by volume usually describes solutions in which both solute and solvent are liquids. The percent by volume is the ratio of the volume of the solute to the volume of the solution expressed as a percent. The volume of the solution is the sum of the volumes of the solute and the solvent. Calculations are similar to those involving percent by mass. volume of solute Percent by volume = 1 fIt' X vo ume 100 0 so u lOn Rubbing alcohol is an aqueous solution of liquid isopropyl alcohol. The label on a typical container, such as the one shown in Figure 15-12, usually states that the rubbing alcohol is 70% isopropyl alcohol. This value is a percent by volume. It tells you that 70 volumes ofisopropyl alcohol are dissolved in every 100 volumes of solution. Because a solution's volume is the sum of the volumes of solute and solvent, there must be 30 volumes of water (solvent) in every 100 volumes of the rubbing alcohol. PROBLEMS 11. What is the percent by volume of ethanol in a solution that contains 35 mL of ethanol dissolved in 115 mL of water 12. If you have 100.0 mL of a 30.0% aqueous solution of ethanol, what volumes of ethanol and water are in the solution 13. What is the percent by volume of isopropyl alcohol in a solution that contains 24 mL of isopropyl alcohol in 1.1 L of water Molarity As you have learned, percent by volume and percent by mass are only two of the commonly used ways to quantitatively describe the concentrations of liquid solutions. One of the most common units of solution concentration is molarity. Molarity (M) is the number of moles of solute dissolved per liter of solution. Molarity also is known as molar concentration. The unit M is read as molar. A liter of solution containing one mole of solute is a 1M solution, which is read as a one molar solution. A liter of solution containing 0.1 mole of solute is a O.lM solution. To calculate a solution's molarity, you must know the volume of the solution and the amount of dissolved solute. Molarit (M) = ~oles of sol~te y liters of solution :. RUBEFACIENT JTOPICAL ANTI""CROE1~l 32 FL OZ (1 an 946 mL For example, suppose you need to calculate the molarity of 100.0 rnL of an aqueous solution containing 0.085 mole of dissolved potassium chloride (KCl). You would first convert the volume of the solution from milliliters to liters using the conversion factor 1 L = 1000 rnL. Figure 15·12 The composition of this isopropyl alcohol is given in percent by volume, which is often expressed as % (v/v). What does each v in the expression 70% (vJv) refer to (l00)Rl::) (lOJo~,)= 0.1000 L Then, to determine the molarity, you would divide the number of moles of solute by the solution volume in liters. 0.085 mol KCI 0.1000 L solution 0.8tmol = 0.85M 464 Chapter 15 <strong>Solutions</strong>
Do the CHEMLAB at the end of this chapter to learn about an experimental technique for determining solution concentration. Calculating Molarity PROBLEM 15-3 A 100.5-mL intravenous (IV) solution contains 5.10 g of glucose (C 6 H 12 0 6 ). What is the molarity of this solution The molar mass of glucose is 180.16 g/mol. II Analyze the Problem You are given the mass of glucose dissolved in a volume of solution. The molarity of the solution is the ratio of moles of solute per liter of solution. Glucose is the solute and water is the solvent. Known II Solve for the Unknown Unknown mass of solute = 5.10 g C 6 H120 6 solution concentration = M molar mass of C6H12 0 6 = 180.16 g/mol volume of solution = 100.5 mL Use the molar mass to calculate the number of moles of C 6 H 12 0 6 . 1 mol C6H1206 (5.10~ ,.. r-l-l-t'l' = 0.0283 mol C 6 H 12 0 6 180.16..y-=6' '12'-'6 ~se ~h~.convers~on factor 1060 L mL to convert the volume of H20 In milliliters to liters. (100.5.J-l'll solution) ( 1060~d= 0.1005 L solution Substitute the known values into the equation for molarity and solve. I 't - moles of solute mo an y - liters of solution . 0.0283 mol C6H 12 0 6 molanty = 0.1005 L solution To prevent dehydration, intravenous (IV) drips are administered to many hospital patients. A solution containing sodium chloride and glucose is commonly used. EI Evaluate the Answer The molarity is asmall value, which is expected because only asmall mass of glucose was dissolved in the solution. The mass of glucose used in the problem contained three significant figures, and therefore, the value of the molarity also has three significant figures. PROBLEMS 14. What is the molarity of an aqueous solution containing 40.0 g of glucose (C 6 H 12 0 6 ) in 1.5 L of solution 15. What is the molarity of a bleach solution containing 9.5 g of NaOCI per liter of bleach 16. Calculate the molarity of 1.60 L of a solution containing 1.55 g of dissolved KBr. 15.2 Solution Concentration 465
Page 1 and 2: Solutions What You'll Learn ~ You w
Page 3 and 4: Solution Formation The intermolecul
Page 5 and 6: Solvation in Aqueous Solutions Why
Page 7 and 8: Heat of solution During the process
Page 9 and 10: As you can see from Table 15-2, the
Page 11 and 12: Using Henry's Law PROBLEM 15-1 If 0
Page 13: Using Percent to Describe Concentra
Page 17 and 18: Diluting solutions In the laborator
Page 19 and 20: Molality and Mole Fraction The volu
Page 21 and 22: Section Ii [olli ative Properties o
Page 23 and 24: Freezing Point Depression The freez
Page 25 and 26: PROBLEMS 33. What are the boiling p
Page 27 and 28: Table 15·6 Types of Colloids Dispe
Page 29 and 30: Figure 15-22 Two examples of the Ty
Page 31 and 32: 6. Pipette 10 mL of blue dye stock
Page 33 and 34: CHAPTER ' STUDY GUIDE Summary 15.1

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