1. A sample of CO2 with a pressure of 76.0 mm Hg in a volume of 0.200 L is compressed so that the new pressure of the gas is 90.0 mm Hg. What is the new volume of the gas? Assume that temperature is constant.
  2. P1 = 76.0 mm Hg; V1 = 0.200 L ; P2 = 90.0 mm Hg ; V2 = ???
  3. P1V1 = P2V2
  4. V2 = P1V1 / P2
  5. V2 = [(76 mm Hg)(.200 L)] / (90 mm Hg) = .169 L
  1. Suppose you have a sample of CO2 in a gas-tight syringe. The gas volume is 38.0 mL at 25˚C. What is the final volume of the gas if you hold the syringe in your hand to raise its temperature to 40˚C?
  2. V1 = 38 mL ; T1 = 25˚C ; T2 = 40˚C ; V2 = ????
  3. T1 = 25 + 273 = 298 K
  4. T2 = 40 + 273 = 313 K
  5. V1 / T1 = V2 / T2
  6. V2 = T2 x (V1 / T1)
  7. V2 = (313 K) x (38 mL / 298 K) = 40 mL
  1. You have a 34.0 L cylinder of helium at a pressure of 110 atm and a temperature of 43˚C. How many balloons can you fill, each with a volume of 26.0 L, on a day when the atmospheric pressure is 600 mm Hg and the temperature is 15˚C?
  2. V1 = 34 L ; P1 = 110 atm ; T1 = 43˚C ; P2 = 600 mm Hg ; T2 = 15˚C ; V2 = ???
  3. P2 = 600 mm Hg x 1 atm / 760 mm Hg = .7895 atm
  4. T1 = 43 + 273 = 316 K
  5. T2 = 15 + 273 = 288 K
  6. General Gas Law  P1V1 / T1 = P2V2 / T2
  7. V2 = (P1V1T2) / (T1P2)
  8. V2 = (110 atm x 34 L x 288 K) / (316 K x .7895 atm) = 4317.57 L
  9. Volume of a balloon is 26L
  10. 4317.57 / 26 = 166.06
  11. The number of balloons that can be filled are 166.
  1. A hot air balloon is filled with 2000 grams of H2. If the temperature of the gas is 32˚C and its pressure is 800 mm Hg, what is the volume of the balloon?
  2. 2000 grams H2 x 1 mol / 1.0079 g = 1984.3238 mol H2
  3. 32 + 273 = 305 K
  4. 800 mm Hg x 1 atm / 760 mm Hg = 1.053 atm
  5. PV = nRT
  6. V = nRT / P
  7. V = [(1984.3238 mol H2)(.08206 (L∙atm)/(mol∙K))(305 K)] / (1.053 atm)
  8. V = 47181 L ≈47200 L
  1. A .306 gram sample of a gaseous compound has a pressure of 400 mm Hg in a volume of 225 mL at 27˚C. What is its molar mass?
  2. Mass = .306 gram ; P = 400 mm Hg ; V = 225 mL ; T = 27˚C
  3. Density of the gas = grams / Liter
  4. V = 225 mL x 1 L / 1000 mL = .225 L
  5. d = .306 grams / .225 L = 1.36 g / L
  6. M = dRT / P
  7. 400 mm Hg x 1 atm / 760 mm Hg = .52634 atm
  8. [(1.36 g / L)(.08206 (L∙atm)/(mol∙K))(300 K)] / (.52632 atm)
  9. M = 63.6 grams / mole
  1. You choose to make some deuterium gas, D2, for use in an experiment and one way to make is to react heavy water, D2O, with the active metal lithium.

2 Li (s) + 2 D2O (l)  2 LiOD (aq) + D2 (g)

Suppose you place .327 grams of Li metal in 20 mL of D2O (d = 1.11 g/mL). What amount of D2 (in moles) can be prepared? If dry D2 gas is captured in a 1325 mL flask at 25˚C, what is the pressure of the gas (in atm)? D2O has a molar mass of 20.03 grams / mole and Li has a molar mass of 6.941 grams / mole.

  1. Moles of Li
  2. 327 g x 1 mole / 6.941 grams = .0471 moles Li
  3. Moles of D2O
  4. 20 mL x 1.11 g / mL = 22.2 grams
  5. 22.2 g x 1 mol / 20.03 g = 1.108 mol D2O
  6. Finding Limiting Reactant
  7. .0471 mol Li x 1 mol D2 / 2 mol Li = .02355 mol D2
  8. 1.108 mol D2O x 1 mol D2 / 2 mol D2O = .554 mol D2O
  9. Li is the limiting reactant
  10. P = nRT / V
  11. T = 25 + 273 = 298 K
  12. V = 1325 mL x 1 L / 1000 mL = 1.325 L
  13. P = [(.02355 mol D2)(.08206 (L∙atm)/(mol∙K))(298 K)] / (1.325 L)
  14. P = .435 atm
  1. Which type of intermolecular force is involved in the compounds below? Rank them according to increasing boiling point.
  2. Ethylene glycol is a polar compound that is capable of hydrogen bonding with water and also other polar molecules. It is therefore dipole – dipole and it also contains LDFs. This is ranked the second highest because it has dipole – dipole forces in addition to hydrogen bonding, but its forces are not as strong as KCl.
  3. Octane is a nonpolar compound. This compound is ranked as having the second lowest boiling point because it only has the LDFs. It is ranked higher than 2,2-dimethylpentae because there is no branching of substituents off the carbon chain. This indicates that there is more intermolecular contact, making it harder to disrupt the IMFs. The compound also has a higher molar mass than 2,2-dimethylpentane. A higher molar mass means more electrons, higher amount of polarizability, and stronger IMFs.
  4. 2,2-dimethylpentane is also non polar. This compound is ranked as having the lowest boiling point because it only has LDFs and there is also branching of substituents off of the carbon chain, making it harder to pack together. When it is harder to pack there is poor intermolecular contact and this enables the IMFs to be easily disrupted.
  5. 3,4-dichlorohexane has dipole - dipole forces because of the polar C-Cl bond. This compound is right in the middle because it has the LDFs like the nonpolar compounds but it also has polar C-Cl bonds. This dipole – dipole forces is strong but they are not as strong as the hydrogen bonding associated with the ethylene glycol and the ionic forces associated with KCl.
  6. Potassium chloride (KCl) is an ionic compound and therefore has ion – ion forces and it also has ion – dipole forces when it is dissolved in water. This is ranked as the compound having the highest boiling point because it has ion – ion forces in addition to ion – dipole forces.
  7. Therefore the rank order from lowest to highest for boiling point is C < B < D < A < E
  1. What is the empirical formula of a unit cell that contains carbon corner atoms, sodium face atoms, and one oxygen body atom?
  2. 8 carbon corner atoms @ 1/8 = 1
  3. 6 sodium face atoms @ ½ = 3
  4. 1 oxygen body atom @ 1 = 1
  5. The empirical formula of this unit cell is CNa3O
  1. As calculated from its x-ray diffraction pattern Tungsten (W) has a cubic unit cell with a side length of 470 pm with a body centered cubit unit cell. Compute the density of W metal in g/cm3. The molar mass of W is 183.84 g/mol.
  2. 470 pm x 1 x 10-12 m / 1 pm = 4.7 x 10-10 m
  3. 4.7 x 10-10 m x 100 cm / 1 m = 4.7 x 10-8 cm
  4. (4.7 x 10-8 cm) 3 = 1.03823 x 10-22 cm3
  5. Body centered cubit unit cell
  6. 8 corners @ 1/8 = 1
  7. 1 body @ 1 = 1
  8. Total number of atoms is 2
  9. 2 atoms x 1 mol / (6.022 x 1023 atoms) = 3.3212 x 10-24 mol
  10. 3.3212 x 10-24 mol x 183.84 grams / mol = 6.106 x 10-22 grams
  11. 6.106 x 10-22 grams / 1.03823 x 10-22 cm3 = 5.88 g/cm3
  1. The air that you breathe (in mole percentage) consists of 1% argon, 21% oxygen and 78% nitrogen. A 2.00 L container is filled with this are at a temperature of 35°C, and the pressure of the container is 1.5 atm. Calculate the partial pressures of each gas in the container.Then find the total pressure and the mole fraction of each gas.
  2. n = PV / RT
  3. T = 35 + 273 = 308 K
  4. n = [(1.5 atm)(2.00 L)] / [(.08206 (L∙atm)/(mol∙K))(308 K)]
  5. n = .119 moles of air.
  6. Calculate the number of moles of each gas
  7. .119 mol air x 1 mol argon / 100 mol air = .00119 mol argon
  8. .119 mol air x 21 mol oxygen / 100 mol air = .0249 mol oxygen
  9. .119 mol air x 78 mol nitrogen / 100 mol air = .0926 mol nitrogen
  10. Find the partial pressure of each gas
  11. Pargon = nRT / V
  12. Pargon = [(.00119 mol)(.08206 (L∙atm)/(mol∙K))(308 K)] /2L
  13. Pargon = 0.015
  14. Poxygen =nRT / V
  15. Poxygen = [(.0249 mol)(.08206 (L∙atm)/(mol∙K))(308 K)] /2L
  16. Poxygen = 0.315
  17. Pnitrogen = nRT / V
  18. Poxygen = [(.0926 mol)(.08206 (L∙atm)/(mol∙K))(308 K)] /2L
  19. Poxygen = 1.17
  20. Find the total pressure
  21. PTOT = Pargon + Poxygen + Pnitrogen
  22. PTOT = .015 + .315 + 1.17 = 1.845
  23. Find the mole fraction of each gas
  24. Pgas = Xgasx PTOT Xgas = Pgas / PTOT
  25. Xargon = .015 / 1.845 = .008
  26. Xoxygen = .315 / 1.845 = .171
  27. Xnitrogen = 1.17 / 1.845 = .634
  1. How does soap work to get rid of grease?
  2. The nonpolar, lipophilic, end interacts with the grease while the polar, hydrophilic, end interacts with the water. The soap disrupts the IMFs in the grease and lowering the surface tension.
  1. In the water phase diagram, why is the slope of the line negative?
  2. The density of liquid water is more than the density of solid water.
  1. Why are covalent solids stronger than ionic solids?
  2. Covalent solids are held together in an extended 3-D network with covalent bonds where as the ionic solids are only held together with ion – ion forces.
  1. Explain the instances where the ideal gas law breaks down.
  2. The ideal gas law breaks down at higher pressure, lower temperature, and higher molar mass.
  1. How many liters of H2 will be required at a temperature of 422 K and 2.4 atm to consume 47 grams of N2? H2 + N2 NH3
  2. Balance the equation
  3. 3H2 + N2 2NH3
  4. Find moles of N2
  5. 47 grams x 1 mol / 28 grams = 1.679 mol N2
  6. Convert moles of N2 to moles of H2
  7. 1.679 mol N2 x 3 mol H2 / 1 mol N2 = 5.036 mol H2
  8. Vhydrogen = nRT / P
  9. Vhydrogen = [(5.036 mol)(.08206 (L∙atm)/(mol∙K))(422)] / 2.4 atm
  10. Vhydrogen = 72.7 L
  1. What is the molecular weight of a gas which diffuses 1/22 as fast as helium?
  2. Graham’s Law
  3. Rate of effusion of gas 1 / rate of effusion of gas 2 = square root (molar mass of gas 2 / molar mass of gas 1)
  4. Let rate 1 = unknown gas = 1
  5. Let rate 2 = helium= 22
  6. Molar mass of He = 4.0026
  7. 1 / 22 = square root (4.0026 / x)
  8. X = 4.0026 / ((1/22)2) = 1937.28