Name: _________________________________________ Date:____________________________ OPTIONAL Homework – do not turn in!! 1.) Which of the following has the greatest polarizabilty? a. Mg+2 or Mg b. Br-1 or I-1 c. F-1 or I-1 Remember that as size increases, so does the electron cloud. A larger electron cloud is more polarizable which means stronger London Dispersion Forces (LDF) 2.) Classify each of the following as a conductor, an insulator, or a semiconductor: a. Phosphorus: insulator (non-metals do not conduct electricity) b. Mercury: conductor (metals DO conduct electricity) c. Germanium: semi-conductor (metalloids have properties of metals and non-metals so it semi conducts!) 3.) State whether the disorder of the system will increase or decrease in each of the following processes: a. A glass vase is shattered: 1 thing turns into many many many pieces. Those pieces are not going to fall together in the shape of the vase, they will go everywhere (think about breaking something glass and then having to clean it up!) there will be much chaos and lack of order with the location of all those pieces!!: DISORDER INCREASES b. Gold is extracted and purified from its ore: a solid-solid mixture is purified into one pure substance (gold) so something that was randomly mixed together is now a single purified item. Chaos and randomness were destroyed in order to turn that vein running through the rock into plain old gold: DISORDER DECREASES (is lost) c. Dry ice (CO2) sublimes: CO2 molecules are locked in place in the solid phase (form). But when sublimation occurs, those molecules in the solid phase turn into the gas phase. Molecules in the gas phase can go wherever they want to – their motion is random so therefore is their disorder. There is much disorder/chaos with a gas: DISORDER INCREASES 4.) Calculate the molarity (FROM TERM 1!!!) of each: review term 1 or look ahead in this chapter notes! a. 42.3 grams of sugar (C12H22O11) in 100.00 mL of solution C: 12 x 12.01 = 144.1 H: 22 x 1.01 = 22.2 O: 11 x 16.00 = 176.0 mm = 342.3 g/mole 42.3 grams of sugar x 1 mole sugar = 0.124 moles sugar 342.3 grams sugar 100.00 mL x 1L = 0.10000 L 1000 mL solution M= 0.124 moles sugar = 1.24 M 0.10000 L solution b. 5.50 grams of LiNO3 in 505 mL of solution Li: 1 x 6.941 = 6.941 N: 1 x 14.01 = 14.01 O: 3 x 16.00 = 48.00 mm = 68.95 g/mole 5.05 g LiNO3 x 1 mole LiNO 3 = 0.0798 moles LiNO3 68.95 grams LiNO 3 M= 505 mL x 1L = 0.505 L 1000 mL 0.0798 moles LiNO 3 = 0.158 M 0.505 L solution 5.) Calculate the molarity of 75.0 mL of 0.250 M NaOH diluted to volume of 0.250 L with water M1V1 = M2V2 (for dilutions!!) M2 = M1 = 0.250 M M2 = ????? 1L = 0.0750 L V1 = 75.0 mL x 1000 mL V2 = 0.250 L M 1 V1 0.250 M x 0.0750 L = = 0.0750 M (3 sig figs!!) 0.0250 L V2 6.) How would you prepare the following aqueous solution: 355 mL of 8.74 x 10-2 M KH2PO4 starting with solid KH2PO4 8.74x10 −2 moles KH 2 PO 4 1L x x 355 mL = 0.0310 moles KH2PO4 1L 1000 mL K: 1 x 39.10 = 39.10 H: 2 x 1.01 = 2.02 P: 1 x 30.97 = 30.97 O: 4 x 16.00 = 64.00 mm = 136.09 g/mole 0.0310 moles KH2PO4 x 136.09 grams KH 2 PO 4 = 4.22 grams KH2PO4 1 mole KH 2 PO 4 First I would take a 355 mL volumetric flask (who cares if there really is one!! It’s the idea of the method that matters!!). I would ½ fill the vol flask with water (solvent), then I would then add the 4.22 grams KH2PO4. I would swirl the solution and then fill to the mark with solvent (water). Finally I would invert the flask several times for thorough mixing. Do not use a beaker. Do not use a graduated cylinder. Another acceptable answer (since the likelihood of finding a vol flask of this size is slim) would be to titrate out 355.00 mL with a buret!! Following the same dissolve and swirl technique. 7.) The Henry’s law constant (kH) for O2 in water at 20oC is 1.28 x 10-3 mol/Latm. How many grams of O2 will dissolve in 2.00 L of H2O when the partial pressure of O2 above the water is 1.00 atm? How many grams of O2 will dissolve in 2.00 L of water when the partial pressure of O2 above the water is 0.502 atm? Henry’s Law: Solubility of the gas = kH x Pgas the solubility of the gas is directly proportional to the pressure of the gas on the liquid. The higher the pressure pushing down on the liquid, the more likely that gas will interact and come in contact with the liquid and get “sucked” in by the IMFs (interactions) of the gas with the liquid S= S= 32.00 grams O 2 1.28 x 10 −3 mole x 2.00 L x 1.00 atm = 2.56 x 10-3 moles of O2 x = 0.0819 grams O2 L atm 1 mole O 2 32.00 grams O 2 1.28 x 10 −3 mole x 2.00 L x 0.502 atm = 1.29 x 10-4 moles of O2 x = 0.0411 grams O2 L atm 1 mole O 2 Notice that when the pressure is almost halved, the amount (in grams!) of O2 is also almost halved! 8.) The partial pressure of CO2 gas above the liquid in a bottle of champagne at 20oC is about 5.5 atm. What is the solubility ( in molecules/L) of CO2 in the champagne where kH CO2 = 3.7 x 10-2 mol/Latm S = kH x Pgas moles CO 2 6.022 x 10 23 molecules CO 2 3.7 x 10 −2 mole x x 5.5 atm = 0.2035 = 1.2 x 1023 CO2 L atm L 1 mole CO 2 molecules L S= Don’t forget that all we are talking about is how much stuff (in this case CO2) is present in a liter of solution (most likely water as the solvent). How much stuff can be in terms of grams, milligrams, kilograms, micrograms, moles, molecules, atoms, ions, etc!! 9.) A solution of isopropanol (C3H7OH) is made by dissolving 0.66 mole of isopropanol in 0.89 mole of water. What is: a. The mole fraction of isopropanol? b. The mass percent of isopropanol? c. The molality of isopropanol? moles x Mole fraction = total moles Xiso = moles isopropanol 0.66 = = moles isopropanol+ moles water (0.66 + 0.89) C3H7OH C: 3 x 12.01= 36.03 H: 8 x 1.01 = 8.08 O: 1 x 16.00 = 16.00 mm = 60.11 grams/mole 0.66 mole iso x Mass percent = molality = H2 O H: 2 x 1.01 = 2.02 O: 1 x 16.00 = 16.00 mm = 18.02 grams/mole 60.11 grams iso = 39.7 grams 1 mole iso 0.89 mole H2O x 18.02 grams water = 16.0 grams 1 mole water grams isopropanol 39.7 grams iso x 100 = x 100 = total grams (39.7 grams iso + 16.0 grams water) moles iso 0.66 moles iso = = 1 kg water 0.016 kg water 16.0 grams water x 0.43 41 molal 1 kg water = 0.016 kg 1000 g water 71%