# Colligative%20Properties%20Answers - 4 The freezing point...

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Colligative Properties 1. List the four colligative properties Vapor pressure lowering, boiling point elevation, freezing point depression, osmotic pressure 2. Solve for the freezing point of benzene if 215 g of CO 2 are added to 1.55 kg of benzene. The K f of benzene is 4.90 C/m and the freezing point of pure benzene is 5.5 Celsius. 215 g * ( 1 mol / 44.01 g) = 4.885 mol m = 4.885 mol / 1.55 kg = 3.1518 mol/kg ΔT f = 3.1518 * 4.90 = 15.444 Celsius Freezing point = 5.5 – 15.44 = -9.9 Celsius 3. Determine the vapor pressure of a solution that has 35.5 g of Na 3 PO 4 dissolved in 935 g of water. The vapor pressure of pure water at the same temperature is 125 torr. Mol Na 3 PO 4 = 35.5 g * (1 mol / 163.94 g) = 0.21654 mol Na 3 PO 4 = 0.8662 mol of ions Mol H 2 O = 935 * (1 mol / 18.02 g) = 51.887 mol Mol fraction (solvent) = 51.887 / (51.887 + 0.8662) = 0.9836 Vapor Pressure (solution) = 0.9836 * 125 = 123 torr
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Unformatted text preview: 4. The freezing point of a solution that is made of MgBr 2 dissolved in 5.95 kg of water is found to be -3.2 Celsius. Determine the mass of MgBr 2 that was dissolved in the water. The K f for water is 1.86. ΔT f = 3.2 Celsius = 1.86 * m m = 1.72 mol ions / kg water 1.72 mol ions / kg water * 5.95 kg water = 10.237 mol ions 10.237 mol ions * ( 1 mol MgBr 2 / 3 mol ions) = 3.412 mol of MgBr 2 3.412 mol MgBr 2 * (184.1 g / mol) = 628 g of MgBr 2 5. How much CO 2 would be needed to add to 1.00 kg of ethanol to make the boiling point 80.0 Celsius. The boiling point of pure ethanol is 78.5 Celsius and the K b for ethanol is 1.22 C/m. Increase in boiling (ΔT b ) = 1.5 Celsius (found by 80 – 78.5) 1.5 = 1.22 * m m = 1.23 mol CO 2 /kg ethanol 1.23 mol CO 2 /kg ethanol * 1.00 kg ethanol = 1.23 mol CO 2 1.23 mol CO 2 * (44.01 g /mol) = 54.1 g...
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## This note was uploaded on 11/29/2011 for the course CHEM 1212 taught by Professor Dockery during the Fall '08 term at Kennesaw.

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