chapter 12 problems

# chapter 12 problems - Solving 2.501000 933.9 2.677 =...

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Chem 112 - Study Guide Solutions 1-11-08 Chapter 12 problems 16,18,22 Problem 16: Calculate the amount of water in grams that must be added to 5.00 g of urea (Molar Mass = 60 g/mole)in the preparation of a 16.2 percent by mass solution. Let x be the number of grams of water that must be added. Solving for x, we find that x = 25.9 g 5.00g 5.00g x g + 100 16.2 Problem 18: Calculate the molality of each of the following aqueous solutions (a) 2.50 M NaCl solution (density of solution = 1.08 g/mL) The molar mass of NaCl is 58.44 g 22.99 35.45 + 58.44 = 2.5058.44 146.1 = 2.50 moles of NaCl has a mass of 146.1 g The mass of one liter of solution is 1080 g The mass of water in the one liter of solution is 933.9 g 1080 146.1 - 933.9 = Let x be the number of moles of NaCl found in one kilogram of water 2.50 933.9 x 1000

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Unformatted text preview: Solving 2.501000 933.9 2.677 = Molality is 2.68 m Problem 22: The concentrated sulfuric acid we use in the laboratory is 98.0 percent H 2 SO 4 mass. Calculate the molality and molarity of the acid solution. The density of the solution is 1.83 g/mL 1 L of this solution must weigh 1830g The mass of the sulfuric acid in the 1L must be 1793 g. .98 1830 1.793 10 3 = The mass of water in the one 1L of solution must be 37 g 1830 1793-37 = The molar mass of sulfuric acid is 98 g/mole 1793 98 18.296 = Since 18.3 moles of sulfuric acid are present in 1L, then the molarity is 18.3 M. Let x be the number of moles of sulfuric acid that would be present in 1000 g of water Solving for x 18.3 37 x 1000 18.31000 37 494.595 = The solution is 494 m...
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## This note was uploaded on 04/07/2008 for the course CHEMISTRY 112 taught by Professor Donato during the Spring '08 term at CofC.

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chapter 12 problems - Solving 2.501000 933.9 2.677 =...

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