| CHEM 128 SAMPLE EXAM 1 KEY |
| 1. Assume a liter of solution. One liter = 896.0 g (from density). Of that mass, 29.89% is ammonia. 0.2989 x 896.0 g = 267.8 g ammonia. Converting g to mol gives you mol/L (M). 267.8 g x (1 mol/17.034 g) = 15.72 M. |
| 2. Molality = mol solute/kg solvent. Molar mass of ethanol = 46.07 g. 1.00 g x (1 mol/46.07 g) = 0.0217 moles 0.0217 mol/0.100 kg = 0l217 m. Mass percent = (1.00 g/101 g) x 100% = 0.990%. Mole fraction: moles of water = 1.00 x 102g x (1 mol/18.02 g) = 5.55 mol. Total moles of solution = 5.55 + 0.0217 = 5.57 mol. Mole fraction = 0.0217/5.57 = 0.00390. |
| 3. All three molecules contain O-H covalent bonds. Acetic acid would be more soluble. Ethanol is a small polar molecule that can form hydrogen bonds. So is acetic acid. Although stearic acid has a polar group that can form hydrogen bonds, most of it consists of a long nonpolar hydrocarbon chain. Therefore stearic acid acts more like a nonpolar substance than a polar substance. |
| 4. increase; decrease; 5. decreases |
| 6. The change in boiling point = kb x m. Calculate the molality of the solution: 4.9 g sucrose/342.31 g/mol = 1.4 x 10-2 mol. m = 1.4 x 10-2 mol/0.175 kg = 0.082 m. Delta Tb = 0.52 °C/m x 0.082m = 0.043°C. This is added on to the normal bp of water, so bp = 100.043°C. |
| 7. pV = nRT. We have the mass of dextran. n = pV/RT. If we calculate n, then the molar mass of dextran = g/n. [(1.47/760)atm x 0.106 L]/(0.0821 L•atm/mol•K x 294 K)= n = 8.49 x 10-6 mol. Molar mass of dextran = 0.582g/8.49 x 10-6 mol = 6.86 x 104 g/mol. |
| 8. Let us assume for starters that CsCl ionizes completely. In that case i = 2. What is the fp depression in that case? Delta Tf = 1.86 °C/m x 0.091 m x 2 = 0.34°C, so fp would be - 0.34°C. However, we are told that the experimental fp depression is 0.302°C. Using this value, calculate i. i = Delta Tf/(m x kf) = 0.302/(0.091 x 1.86) = 1.8. Based on this calculation, we can safely say that CsCl is mostly ionized, since 1.8 is closer to 2 than to 1. |
| 9. See answer key for Assignment 1. |
| 10. This reaction is first order in A, second order in B, and third order overall. |
| 11. See answer key for Assignment 3. |
| 12. a.
The reaction is first order. b. Using a form of the first order integrated rate law, ln ([A]0/[A]) = kt, ln(0.24M/0.12M)/4.0 x 10-4s-1 = 1.7 x 103 s. Notice that the second concentration is half the first concentration, so what you have found is the half-life, t1/2. c. Rate = k[ethane], or rate = 4.0 x 10-4s-1 x [ethane] |
| 13. and 14. See Assignment 5 Key. |
| 15. As the temperature is raised, a greater fraction of reactant collisions has energy equal to or greater than the activation energy, Ea. (The rate constant k is dependent on p x Z x e-Ea/RT, where e-Ea/RT is the fraction f of collisions with energy equal to or greater than Ea. When T increases, the exponent becomes a smaller negative exponent, so f becomes greater.) |
| 16. A catalyst lowers the activation energy for a reaction by changing the reaction mechanism or pathway. Lower activation energy is associated with increased reaction rate. |