# Ch11 - 11 Solutions and Their Properties 11.1 Toluene is nonpolar and is insoluble in water Br 2 is nonpolar but because of its size is polarizable

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Unformatted text preview: 11 Solutions and Their Properties 11.1 Toluene is nonpolar and is insoluble in water. Br 2 is nonpolar but because of its size is polarizable and is soluble in water. KBr is an ionic compound and is very soluble in water. toluene &amp;lt; Br 2 &amp;lt; KBr (solubility in H 2 O) 11.2 (a) Na + has the larger (more negative) hydration energy because the Na + ion is smaller than the Cs + ion and water molecules can approach more closely and bind more tightly to the Na + ion. (b) Ba 2+ has the larger (more negative) hydration energy because of its higher charge. 11.3 NaCl, 58.44 amu; 1.00 mol NaCl = 58.44 g 1.00 L H 2 O = 1000 mL = 1000 g (assuming a density of 1.00 g/mL) mass % NaCl = 100% x g 58.44 + g 1000 g 58.44 = 5.52 mass % 11.4 ppm = solution of mass total CO of mass 2 x 10 6 ppm total mass of solution = density x volume = (1.3 g/L)(1.0 L) = 1.3 g 35 ppm = g 1.3 CO of mass 2 x 10 6 ppm mass of CO 2 = ppm 10 g) ppm)(1.3 (35 6 = 4.6 x 10 5 g CO 2 11.5 Assume 1.00 L of sea water. mass of 1.00 L = (1000 mL)(1.025 g/mL) = 1025 g 100% x g 1025 NaCl mass = 3.50 mass %; mass NaCl = 100 3.50 x g 1025 = 35.88 g There are 35.88 g NaCl per 1.00 L of solution. M = L 1.00 NaCl g 58.44 NaCl mol 1 x NaCl g 35.88 = 0.614 M 11.6 C 27 H 46 O, 386.7 amu; CHCl 3 , 119.4 amu; 40.0 g x g 1000 kg 1 = 0.0400 kg molality = kg 0.0400 g 386.7 mol 1 x g 0.385 = CHCl kg O H C mol 3 46 27 = 0.0249 mol/kg = 0.0249 m 255 CHCl mol + O H C mol O H C mol = X 3 46 27 46 27 O H C 46 27 X O H C 46 27 = g 119.4 mol 1 x g 40.0 + g 386.7 mol 1 x g 0.385 g 386.7 mol 1 x g 0.385 = 2.96 x 10 3 11.7 CH 3 CO 2 Na, 82.03 amu kg H 2 O = Na CO CH mol 0.500 O H kg 1 Na) CO CH mol (0.150 2 3 2 2 3 = 0.300 kg H 2 O mass CH 3 CO 2 Na = 0.150 mol CH 3 CO 2 Na x Na CO CH mol 1 Na CO CH g 82.03 2 3 2 3 = 12.3 g CH 3 CO 2 Na mass of solution needed = 300 g + 12.3 g = 312 g 11.8 Assume you have a solution with 1.000 kg (1000 g) of H 2 O. If this solution is 0.258 m , then it must also contain 0.258 mol glucose. mass of glucose = 0.258 mol x mol 1 g 180.2 = 46.5 g glucose mass of solution = 1000 g + 46.5 g = 1046.5 g density = 1.0173 g/mL volume of solution = g 1.0173 mL 1 x g 1046.5 = 1028.7 mL volume = 1028.7 mL x mL 1000 L 1 = 1.029 L; molarity = L 1.029 mol 0.258 = 0.251 M 11.9 Assume 1.00 L of solution. mass of 1.00 L = (1.0042 g/mL)(1000 mL) = 1004.2 g of solution 0.500 mol CH 3 CO 2 H x H CO CH mol 1 H CO CH g 60.05 2 3 2 3 = 30.02 g CH 3 CO 2 H 1004.2 g 30.02 g = 974.2 g = 0.9742 kg of H 2 O; molality = kg 0.9742 mol 0.500 = 0.513 m 11.10 Assume you have 100.0 g of seawater....
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## This note was uploaded on 08/27/2008 for the course CHEM 1280 taught by Professor Schmidt during the Fall '07 term at Toledo.

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Ch11 - 11 Solutions and Their Properties 11.1 Toluene is nonpolar and is insoluble in water Br 2 is nonpolar but because of its size is polarizable

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