ch14 - Chapter 14: Solutions 1 I.More Concentration Units...

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1 Chapter 14: Solutions
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2 I.More Concentration Units A.Molar concentration B.Mole fraction and mole percent C.Weight fraction and weight percent D.Molal concentration II.The Solution Process A.“Like dissolves like” B.Heats of solution C.Factors affecting solubility III.Colligative Properties A.Vapor pressure: Raoult's law B.Boiling point elevation and freezing point depression C.Molar masses from colligative properties D.Osmotic pressure E.Solutions of electrolytes
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3 molarity, M = moles of solute liter of solution mole fraction of compound A in a mixture, X A = (no units) mole percent of A = X A · 100 1.0 mol of NaCl is dissolved in 4.0 mol of water: X NaCl = mol % NaCl = X water = mol % H 2 O = n A n total I.More Concentration Units A.Molar concentration A.Mole fraction and mole percent
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4 w A = % (w/w) = w A · 100 Vinegar is 5.00 % (w/w) acetic acid in water. How many grams of acetic acid are in one cup (237 mL) of vinegar? (d vinegar = 1.0055 g/cm 3 ) mass of A total mass molality, m = moles of solute kg of solvent The solubility of NaCl at 0ºC is 35.7 g/100 g of water. What is the molality of this solution? (fw NaCl 58.44) I.More Concentration Units A.Weight fraction and weight percent A.Molal concentration
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5 I.More Concentration Units A.Molal concentration Vinegar is 5.00 % (w/w) acetic acid in water. The density of vinegar is 1.0055 g/mL. What are X, m , and M ? (mw HC 2 H 3 O 2 60.05)
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6 Antifreeze is typically a 50/50 (v/v) mixture of ethylene glycol (C 2 H 6 O 2 ) in water. What are m and M ? (mw C 2 H 6 O 2 62.06) d (C 2 H 6 O 2 ) = 1.1155 g/mL d (H 2 O) = 1.0000 g/mL d (50/50) = 1.069 g/mL I.More Concentration Units A.Molal concentration
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7 O H H O H O H H CH 3 CH 3 OH( l ) + H 2 O( l ) miscible H-bonds in pure substances are replaced by H-bonds in solution H-bonding H-bonding CCl 4 ( l ) + C 6 H 6 ( l ) miscible London London London forces in pure substances are replaced by London forces in solution but CCl 4 ( l ) + H 2 O( l
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ch14 - Chapter 14: Solutions 1 I.More Concentration Units...

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