Physics Solution Manual for 1100 and 2101

Physics Solution Manual for 1100 and 2101

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Unformatted text preview: 3 Mass of 2 nitrogen atoms Mass of 5 oxygen atoms b. The mass per mole of aspartame is 294.307 g/mol. The number of aspartame molecules per mole is Avogadro’s number, or 6.022 × 1023 mol–1. Therefore, the mass of one aspartame molecule (in kg) is 294.307 g/mol 1 kg –25 = 4.887 ×10 kg 23 −1 6.022 ×10 mol 1000 g ______________________________________________________________________________ 4. REASONING The mass of one of its atoms (in atomic mass units) has the same numerical value as the element’s mass per mole (in units of g/mol). Atomic mass units can be converted into kilograms using the fact that 1 u = 1.6605 × 10–27 kg. Dividing the mass of the sample by the mass per mole gives the number of moles of atoms in the sample. 724 THE IDEAL GAS LAW AND KINETIC THEORY SOLUTION a. The mass per mole is 196.967 g/mol. Since the mass of one of its atoms (in atomic mass units) has the same numerical value as the mass per mole, the mass of a single atom is m = 196.967 u . b. To convert the mass from atomic mass units to kilograms, we use the conversion factor of 1 u = 1.6605 × 10–27 kg: 1.6605 × 10−27 kg −25 m = (196.967 u ) = 3.2706 × 10 kg 1u c. The number n of moles of atoms is equal to the mass m divided by the mass per mole: 285 g m = = 1.45 mol Mass per mole 196.967 g /mol ______________________________________________________________________________ n= 5. SSM REASONING The mass (in grams) of the active ingredient in the standard dosage is the number of molecules in the dosage times the mass per molecule (in grams per molecule). The mass per molecule can be obtained by dividing the molecular mass (in grams per mole) by Avogadro’s number. The molecular mass is the sum of the atomic masses of the molecule’s atomic constituents. SOLUTION Using N to denote the number of molecules in the standard dosage and mmolecule to denote the mass of one molecule, the mass (in grams) of the active ingredient in the standard dosage can be written as follows: m = Nmmolecule Using M to denote the molecular mass (in grams per mole) and recognizing that M mmolecule = , where NA is Avogadro’s number and is the number of molecules per mole, NA we have M m = Nmmolecule = N N A M (in grams per mole) is equal to the molecular mass in atomic mass units, and we can obtain this quantity by referring to the periodic table on the inside of the back cover of the text to find the molecular masses of the constituent atoms in the active ingredient. Thus, we have Chapter 14 Problems 725 Molecular mass = 22 (12.011 u ) + 23 (1.00794 u ) + 1( 35.453 u ) + 2 (14.0067 u ) + 2 (15.9994 u ) 14 244 14 244 14243 14 244 14 244 4 3 4 3 4 3 4 3 Carbon Hydrogen Chlorine Nitrogen Oxygen = 382.89 u The mass of the active ingredient in the standard dosage is M m= N NA 6. 382.89 g/mol 19 −2 = 1.572 × 10 molecules = 1.00 × 10 g 23 6.022 × 10 molecules/mol ( ) REASONING AND SOLUTION The number n of moles contained in a sample is equal to the number N of atoms in the sample divided by the number NA of atoms per mole (Avogadro’s number): 23 N 30.1×10 = = 5.00 mol n= N A 6.022×1023 mol −1 Since the sample has a mass of 135 g, the mass per mole is 135 g = 27.0 g/mol 5.00 mol The mass per mole (in g/mol) of a substance has the same numerical value as the atomic mass of the substance. Therefore, the atomic mass is 27.0 u. The periodic table of the elements reveals that the unknown element is aluminum . ______________________________________________________________________________ 7. REASONING a. The number n of moles of water is equal to the mass m of water divided by the mass per mole: n = m/(Mass per mole). The mass per mole (in g/mol) of water has the same numerical value as its molecular mass (which we can determine). According to Equation 4.5, the total mass of the runner is equal to her weight W divided by the magnitude g of the acceleration due to gravity, or W/g. Since 71% of the runner’s total mass is water, the mass of water is m = (0.71)W/g. b. The number N of water molecules is the product of the number n of moles and Avogadro’s number NA (which is the number of molecules per mole), or N = n NA . SOLUTION a. Starting with n = m/(Mass per mole) and substituting in the relation m = (0.71)W/g, we have 726 THE IDEAL GAS LAW AND KINETIC THEORY (0.71)W m g n= = Mass per mole Mass per mole (1) The molecular mass of water (H2O) is 2 (1.00794 u) + 15.9994 u = 18.0153 u. The mass per mole is then 18.0153 g/mol. However, we need to convert this value into kilograms per mole for use in Equation (1). This is because the value for the weight W in Equation (1) is given in newtons (N), which is a SI unit. The SI unit for mass is the kilogram (kg), not the gram (g). Converting the mass per mole value to kilograms per mole gives Mass per mole = 18.0153 g 1 kg g = 18.0153 mol mol 103 g Substituting this relation into Equation 1 gives (0.71)W (0.71) ( 580 N ) g 9.80 m/s 2 n= = = 2.3 × 103 mol Mass per mole g 1 kg 18.0153 mol 103 g b. The number of water molecules in the runner’s body is N = n NA = ( 2.3 ×...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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