# More%20Colligative%20Answers - 2 S dissolved in a solution...

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More Colligative Properties 1. Given the freezing point of camphor is 178.75 and k f = 37.7 C/m, what is the boiling point if 75 g of CF 4 is added to 625 g of camphor? ΔT f = 37.7 * (0.85218 mol / 0.625 kg) = 51.40 T f = 178.75 - 51.40 = 127.35 2. The vapor pressure of a sample of benzene (C 6 H 6 ) is 125 torr at a certain temperature. When an unknown amount of C 6 H 12 O is added to 1050 g of benzene, the vapor pressure becomes 113 torr. Solve for the mass of C 6 H 12 O that is added. Solute: C 6 H 12 O, unknown amount Solvent: C 6 H 6 , 1050 g = 13.44 mol 113 = χ solv 125 χ solv = 0.904 = 13.44 / (13.44 + mol(C 6 H 12 O)) 0.904(13.44 + mol(C 6 H 12 O)) = 13.44 (13.44 + mol(C 6 H 12 O)) = 14.867 mol(C 6 H 12 O) = 1.427 mol = 142.98 grams

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3. Solve for the osmotic pressure of a solution that has 25 g of Na
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Unformatted text preview: 2 S dissolved in a solution of 1.50 L. 25 g Na 2 S = 0.3203 moles of Na 2 S = 0.9609 moles of ions M = 0.9609 moles / 1.50 L = 0.6406 M **Oops, I forgot to give a temperature, let's go with 25 Celsius** = 0.6406 * 0.0821 * 298.15 = 15.68 atm 4. Find the boiling point of the solution in question 3, given that the solvent is water, the density of the solution is 1.05 g/mL and the k b 0.515 C/m. solute, Na 2 S has 0.9609 moles of ions present solvent, water, unknown mass solution, 1.50 L = 1500 mL * 1.05 g/mL = 1575 grams Mass Solvent = Mass Solution - Mass Solute = 1575 g - 25 g = 1550 g molality = 0.9609 moles / 1.550 kg = 0.6199 T b = 0.515 * 0.6199 = 0.319 T b = 100.3 Celsius...
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## More%20Colligative%20Answers - 2 S dissolved in a solution...

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