1036 HW 3 _sol_

# 1036 HW 3 _sol_ - General Chemistry 1036 Homework 3...

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General Chemistry 1036 Homework 3 Solutions 1. Consider a solution of aqueous (NH 4 ) 2 CO 3 prepared by mixing 12.00 g of (NH 4 ) 2 CO 3 with 125 g of water to make 127 mL of solution. a. Calculate the percent by mass of (NH 4 ) 2 CO 3 Mass of solute = 12.00 g (NH 4 ) 2 CO 3 Mass of solvent = 125 g water Mass of solution = 137 g (solute + solvent) % = g solute g solution (100) = 12.00 g 137 g (100) = 8.76% b. Calculate the mole fraction of (NH 4 ) 2 CO 3 Moles (NH 4 ) 2 CO 3 : 12.00 g (NH 4 ) 2 CO 3 x 1 mol (NH 4 ) 2 CO 3 96.1 g = 0.125 mol (NH 4 ) 2 CO 3 Moles of H 2 O: 125 g H 2 O x 1 mol H 2 O 18.0 g = 6.94 mol H 2 O 42 3 3 (NH ) CO 3 2 moles (NH ) CO 0.125 moles X = = = 0.0177 moles (NH ) CO + moles H O 0.125 + 6.94 moles c. Calculate the molarity of the solution. 3 mol (NH ) CO M = L of solution mol of (NH 4 ) 2 CO 3 = 0.125 mol (from part b) L of solution = 127 mL = 0.127 L (given in original problem) 0.125 mol M = = 0.984 M 0.127 L d. Calculate the molality of the solution 3 2 mol (NH ) CO m = kg solvent (H O) 0.125 mol (NH 4 ) 2 CO 3 (from part b) 125 g = 0.125 kg H 2 O (given in original problem) 3 2 0.125 mol (NH ) CO m = = 1.00 m 0.125 kg H O

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2. A 10.0 m solution of NaOH has a density of 1.30 g/mL at room temperature. For the solution calculate: a. the mole fraction of NaOH You are given the molality of the solution; 10.0 m 2 mol NaOH m = kg H O Assume an initial amount of solution; for molality, it is easiest to assume an amount of solvent of 1000 g (1 kg) 2 moles of NaOH 10.0 m = 1 kg H O moles of NaOH = 10.0 moles in 1000g (1 kg) of water. For mole fraction, you need moles of NaOH (10.0 moles) and moles of water: 22 2 1000 g H O 1 mol H O x = 55.6 mol H O 18.0 g NaOH 2 moles NaOH 10.0 mol X = = = 0.152 moles NaOH + moles H O 10.0 + 55.6 mol b. its molarity For molarity, you need the moles of NaOH (10.0 mol) and the volume of the solution: How much total solution do you have? You assumed an amount of solvent of 1000g + you have 10.0 moles of NaOH; g of NaOH: 10.0 mol NaOH x 40.0 g NaOH 1 mol NaOH = 400 g NaOH So, you have 1000 g of water and 400 g of NaOH for a total mass of solution of 1000 + 400 = 1400 g of solution.
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1036 HW 3 _sol_ - General Chemistry 1036 Homework 3...

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