Gen Chem II Chapter 13

Gen Chem II Chapter 13 - CHEM 1312 Homework Assignments...

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CHEM 1312 Homework Assignments ANSWER KEY Section 1 [Chemistry (2 nd ed): Burdge] NOTE: Questions that do not have answers are review questions [opened ended or definition style problems]—students should be able to find the answer in the text portion of the perspective chapter. All problems which require a step-by-step solved solution are provided below. See your instructor if you need further help. Chapter 13: 13.10 Strong hydrogen bonding (dipole-dipole attraction) is the principal intermolecular attraction in liquid ethanol, but in liquid cyclohexane the intermolecular forces are dispersion forces because cyclohexane is nonpolar. Cyclohexane cannot form hydrogen bonds with ethanol, and therefore cannot attract ethanol molecules strongly enough to form a solution. __________________________________________________________________________________________________ 13.12 The longer the C C chain, the more the molecule "looks like" a hydrocarbon and the less significant the OH group becomes. Hence, as the C C chain length increases, the molecule becomes less polar. Since “like dissolves like”, as the molecules become more nonpolar, the solubility in polar water decreases. The OH group of the alcohols can form strong hydrogen bonds with water molecules, but this property diminishes in importance as the chain length increases. __________________________________________________________________________________________________ 13.16 (a) The percent by mass is defined by Equation 13.2 as mass of solute percent by mass of solute 100% mass of solute + mass of solvent = ! Substituting in the percent by mass of solute and the mass of solute, we can solve for the mass of solvent (water). 5.00 g urea 16.2% 100% 5.00 g urea + mass of water = ! (0.162)(mass of water) = 5.00 g (0.162)(5.00g) mass of water = 25.9 g (b) Similar to part (a), 2 2 26.2 g MgCl 1.5% 100% 26.2 g MgCl + mass of water = ! mass of water = 1.72 × 10 3 g __________________________________________________________________________________________________ 13.18 moles of solute molality mass of solvent (kg) = (a) 1.08 g mass of 1 L soln 1000 mL 1080 g 1 mL = ! =
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kg 931 . 0 NaCl mol 1 NaCl g 8.44 5 NaCl mol 55 . 2 g 1080 water of mass = ! " # $ % & ( = molality = = O H kg 0.931 NaCl mol .55 2 2 2.74 m (b) 100 g of the solution contains 45.2 g KBr and 54.8 g H 2 O. KBr mol 380 . 0 KBr g 119.0 KBr mol 1 KBr g 2 . 5 4 KBr of mol = ! = O H kg 0548 . 0 g 1000 kg 1 O H g 8 . 4 5 kg) (in O H of mass 2 2 2 = ! = molality = 2 0.380 mol KBr 0.0548 kg H O = 6.93 m __________________________________________________________________________________________________ 13.22 (a) Convert mass percent to molality, where moles of solute molality mass of solvent (kg) = . We first convert 98.0 g H 2 SO 4 to moles of H 2 SO 4 using its molar mass, then we convert 2.0 g of H 2 O to units of kilograms. 2 4 2 4 2 4 2 4 1 mol H SO 98.0 g H SO 0.999 mol H SO 98.09 g H SO ! = 3 2 2 1 kg 2.0 g H O 2.0 10 kg H O 1000 g ! " = " Lastly, we divide moles of solute by mass of solvent in kg to calculate the molality of the solution. 3 mol of solute 0.999 mol kg of solvent 2.0 10 kg ! = = = " 2 5.0 10 m m " (b) Convert molality to molarity. From part (a), we know the moles of solute (0.999 mole H 2 SO 4 ) and the mass of the solution (100.0 g). To solve for molarity, we need the volume of the solution, which we can calculate from its mass and density. First, we use the solution density as a conversion factor to convert to volume of solution.
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Gen Chem II Chapter 13 - CHEM 1312 Homework Assignments...

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