Answers to Colligative Property Questions 1. Calculate the molality of a solution of an unknown element in carbon disulfide if 35.5 g of the element are dissolved in 100.0 g of CS 2 , producing a sln with a boiling point of 49.480 C . m = 3.25/2.35 = 1.38 m 2. For the above sln, calculate the molecular weight (or molar mass) of the unknown element. What is the element? moles = (1.38 mol/kg)(0.1000kg)= 0.138 mol MW = 35.5g/0.138mol = 257 g/mol or 257 amu. Element is Fermium. 3. Determine the molecular weight of a weak acid if a sln contains 30.0g of the acid per kg of water. This sln freezes at -0.930 C. m = 0.93/1.853 = 0.502mol/kg As is 30.0 g /kg, then have: 30.0g/0.502mol = 59.8g/mol 4. The unknown acid above is acetic acid, CH 3 CO 2 H. Calculate your %-error. %-Error = 100%(59.8 – 60.0)/60.0 = 0.333% 5. What adjustment might you have to make to your calculation if the acid had been hydrochloric acid or nitric acid (strong acids, what do they do)? Strong acids ionize in water to give more than 1 particle, so the van’t hoff factor i would
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This note was uploaded on 06/25/2008 for the course CHEM 400-401 taught by Professor Dr.samples during the Fall '06 term at American River.