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ch116-ch11hw - CH116 General Chemistry 2 Solutions to HW...

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Prepared by Prof. Anju Sharma for students in General Chemistry 2 1 CH116 - General Chemistry 2 - Solutions to HW Problems - Chapter 11 Chapter 11: Assigned Problems 32, 36, 40, 52, 58, 60, 64, 66, 70, 78, 80, 86, 94, 96, 98, 100, 110, 120, 121. 32. In lab you need to prepare at least 100 mL of each of the following solutions. Explain how you would proceed using the given information? (a) 2.0 m KCl in water (density of H 2 O = 1.00 g/cm 3 ) 2.0 m KCl means 2 mol KCl in 1 kg of water. Since the density of water is 1.00 g/cm 3 (1 cm 3 = 1 mL); 1 kg of water equals 1 L. We need 100 mL of solution. Assuming that the volume of solution does not change upon the addition of solid, we need 100 mL (100 g of water) O H mL 1000 KCl mol 2 2 x KCl mol 1 KCl g 74.55 x 100 mL H 2 O = 14.91 g KCl must be dissolved in 100 mL of water. (b) 15% NaOH by mass in water (d = 1.00 g/cm 3 ) In every 100 g (100 mL) of a 15% by mass solution of NaOH: mass of NaOH = 15 g; mass of water = 100 – 15 = 85 g Assuming volume of the solution will not increase by adding NaOH to water, this will not give 100 mL of solution needed. Let us calculate the amount of NaOH for 100 g of water. 100 g H 2 O x O H g 85 NaOH g 15 2 = 17.64705882 g of NaOH must be dissolved in 100 g (100 mL) of water. (c) 25% NaOH by mass in CH 3 OH (d = 0.79 g/cm3) Similar to (b) above 25 g of NaOH and 75 g of CH 3 OH make 100 g of solution. Again, assuming the addition of NaOH will not make for a significant increase in volume of the solution, let us calculate for 100 mL of CH3OH. 100 mL CH 3 OH x OH CH mL 1 OH CH g 0.79 3 3 x OH CH g 75 NaOH g 25 3 = 26.33333333 g NaOH must be dissolved in 100 mL of methanol, CH 3 OH. (d) 0.10 mole fraction of C 6 H 12 O 6 in water (d = 1.00 g/cm 3 ) mole fraction of glucose = 0.10; mole fraction of water = 0.90 molar mass of glucose C 6 H 12 O 6 = 180.156 g/mol; molar mass of water H 2 O = 18.016 g/mol Method 1: Assuming we need 100 ml (100 g) of water to make sure we have at least 100 mL of solution: 100 g H 2 O x O H g 18.016 O H mol 1 2 2 x O H mol 9 O H C mol 1 2 6 12 6 = 0.6167357411 mol glucose x 6 12 6 6 12 6 O H C mol 1 O H C g 180.156 = 111.1086442 g of glucose must be dissolved in 100 ml of water. Method 2 : 100 g H 2 O x O H g 18.016 O H mol 1 2 2 = 5.55555555 mol H 2 O (glucose) = 0.10 = O H mol 5.56 + glucose mol glucose mol 2 x x ; 0.10( x + 5.56) = x 0.90 x = 0.556; x = 0.61728395 mol glucose x 6 12 6 6 12 6 O H C mol 1 O H C g 180.156 = 111.2074073 g of glucose.
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Prepared by Prof. Anju Sharma for students in General Chemistry 2 2 36. Calculate the molarity and mole fraction of acetone in a 1.00 m solution of acetone (CH 3 COCH 3 ) in ethanol (C 2 H 5 OH). Density of acetone = 0.788 g/cm 3 ; density of ethanol = 0.789 g/cm 3 ). Assume that the volumes of acetone and ethanol add. 1.00 m acetone in ethanol = ethanol kg 1 acetone mol 1 molar mass of acetone = 58.078 g/mol; molar mass of ethanol = 46.068 g/mol 1 mol acetone x acetone mol 1 acetone g 58.078 x acetone g 0.788 acetone mL 1 = 73.70304569 mL acetone 1000 g ethanol x ethanol g 0.789 ethanol mL 1 = 1267.427123 mL ethanol Total volume of solution = 73.70304569 + 1267.427123 mL = 1341.130169 mL = 1.341 L molarity = L) (in solution of volume acetone mol = solution L 1.341 acetone mol 1 = 0.7456397771 M mol of acetone = 1; mol of ethanol = 1000 g ethanol x ethanol g 46.068 ethanol mol 1 = 21.70704176 mol ethanol (acetone) = ethanol mol + acetone mol acetone mol = 21.71 + 1 1 = 0.044039202 40. (a) Use the following data to calculate the enthalpy of hydration for calcium chloride and calcium iodide.
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