            Main   Tutorials   Organic Chemistry   Practice Tests   Online Quizzes   Reference Tools   Solutions Practice Questions A solution is prepared by mixing 1.0 grams of benzene (C6H6) in 100 g of water to create a solution total volume of 100 ml. Calculate the molarity, mass percent, mole fraction, and molality of benzene in the solution. Which ion would you expect to be more heavily hydrated? K+, Ca2+ Na+, K+ Cu2+, Cu+ Assuming the CO2 partial pressure in air above a lake at sea level is 4.0 x 10-4 atm, what is the equilibrium concentration of CO2 in the lake at 25oC? (Henry’s law constant is 32 L . atm / mole) A cocktail was prepared by mixing 20.0g ethanol to 150 g water at 25oC. Pure water has a vapor pressure of 23.76 torr. What would be the new vapor pressure? 10 grams of salt (NaCl) is added to 100 mL of water. What are the new freezing and boiling points? (Kb = .51 oC kg/mol, Kf = .1.86 oC kg/mol) Calculate the pressure needed to prevent osmosis when 10.0 g CaCl2 is added to 100 mL of water? Answers 1. molarity (1.0 grams benzene) / (78 g/mol) = .0128 moles benzene (0.128 moles0 / (0.100 L) = 1.28 Moles/Liter mass percent total mass = 1.0 g benzene + 100 g water = 101 g 1.0 grams benzene / (total mass) 1.0 g benezine / 101 g * 100 = .99 percent benzene mole fraction (1.0 grams benzene) / (78 g/mol) = .0128 moles benzene (100 g water) / (16 g/mol) = 6.25 moles water (.0128 moles benzene) / (.0128 moles benzene + 6.25 moles water) = .20 % benzene molality (1.0 grams benzene) / (78 g/mol) = .0128 moles benzene (.0128 moles benzene) / (.1 kg water) = 1.28 mole/Kg 2. Ca2+ should be more heavily hydrated because there is a greater attractive force due to the extra charge making D H3 more negative. D H1 and D H2 are going to be similar because these are similarly sized ions. Na+ should be more heavily hydrated because it is the smaller ion, which will decrease D H1 and D H2. Cu2+ should be more heavily hydrated because of the same reasons as in part a. 3. PCO2 = kCO2CCO2 CCO2 = PCO2 / kCO2 CCO2 = 4.0 E-4 atm / 32 L atm/mol CCO2 = 1.2 E-5 mol/L 4. P = c solvent Po 20 g ethanol / 46 g/mol = .44 mol ethanol 150 g water / 16 g.mol = 9.4 mol water c solvent = 9.4 mol water / (9.4 mol water + .44 mol ethanol) c solvent = .96 P = 0.96 * 23.76 mm Hg = 23 mm Hg 5. msolute = moles NaCl / Kg water msolute = (10 g / 58.5 g/mol) / .10 kg water msolute = 1.7 mol/kg Tb = Tob + Kb msolute Tb = 100 + .51* 1.7 * 2 Tb = 102 oC Tf = Tof - Kb msolute Tf = 0 – 1.86* 1.7 * 2 Tf = -6.3 oC Note that i in this case is 2 6. 1.0 g CaCl2/ 111.0 g/mol = .00900 mol CaCl2 .00900 mol CaCl2 / .100 L = .0900 M CaCl2 p = iMRT p = 3*.0900*.08206 * 298 p = 6.60 atm news | about us | contact us tutorials index | organic chemistry | practice tests | online quizzes | reference tools site copyright (c) 2002-2013 Learn Chem   Subscribe to our low volume newsletter to receive up-to-date information about the CHEM SITE       