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3 School of Science and Engineering (SSE) Spring 2017 General Chemistry II (CHE 1402) Homework 3 - CORRECTION I 1. Which of the following choices has the compounds correctly arranged in order of increasing solubility in water? (Justify your answer) A) CH3OH < CH4 < LiF B) LiF < NaNO3 < CHCl3 C) CCl4 < CHCl3 < NaNO3 D) CH3OH < CCl4 < CHCl3 E) CH4 < NaNO3 < CHCl3 CCl4 is non polar and consequently has a very low solubility in water CHCl3 is polar and soluble in water via London dispersion and dipole-dipole interaction forces. NaNO3 is an ionic compound and more soluble in water (ion-dipole interactions) Answer is C. Note: “like dissolves like". 2. At 25 °C the vapor pressure of pure benzene (C6H6) is 95.1 torr and that of pure toluene (C7H8) is 28.4 torr. A solution, is made by mixing equal number of moles of benzene and toluene, has a vapor pressure of 49.4 torr at 25 °C. a. If the solution exhibits ideal behavior, what would be its vapor pressure? b. Explain in terms of intermolecular interactions whether the mixing of benzene and toluene is endothermic or exothermic. a. If the solution exhibits an ideal behavior, then Raoult’s law can be written as: PTOTAL = PC6H6 + PC7 H8 = X C6H6 PCo6H6 + X C7 H8 PCo7 H8 Since the solution is made by mixing equal number of moles of benzene and toluene then n C6 H 6 n C6 H 6 1 X C7 H8 X C6 H 6 0.5 n C6 H 6 n C7 H8 2 n C6 H 6 2 PTOTAL = X C6H6 PCo6H6 + X C7 H8 PCo7 H8 X C6H6 PCo6H6 PCo7 H8 0.5 95.1 28.4 61.7 torr c. The real vapor pressure is 49.5 torr which is lower than the 61.7 torr (calculated ideal vapor pressure); this implies that this case corresponds to a negative deviation from the ideal solution (Raoult’s law) and consequently Hsolute-solvent > (Hsolute-solute + Hsolvent-solvent). This means that the solute-solvent interactions are much stronger than solute-solute interactions and the solvent-solvent interactions. Consequently Hsolution < 0 and the process is exothermic. II Sodium metal Na(s) dissolves in liquid mercury (Hg(l)) to form a solution called sodium amalgam. The density of Na(s) is 0.97g/cm3. A solution is made by dissolving 4.85 g of Na(s) in 20 cm3 of Hg(l); if the density of the solution is 11.07 g/cm3 determine: a. the molality of Na(s) in the solution b. the molarity of Na(s) in the solution a. The molality, m, is defined as: n solute in mol n in mol massNa m so m Na Na massHg in kg M Na massHg in kg masssolvent in kg massHg masssolution massNa d solution Vsolution massNa Assuming that the volume is additive, massNa 4.85 3 20 Vsolution VHg VNa VHg 25 cm 25 mL 0.97 d Na massHg d solution Vsolution massNa 11.07 25 4.85 271.9 g 0.2719 kg m Na massNa 4.85 0.776 mol / kg 0.776 m M Na massHg 23.0 0.2719 n solute in mol Vsolution in L 4.85 8.43 mol / L 8.43 M 23.0 25 10 3 c. The molarity is defined as: molarity so molarity Na massNa M Na Vsolution masssolvent in kg Note: molarity molality Vsolution in L massHg 0.2719 0.776 molarity Na mNa 8.43M 3 25 10 Vsolution III 1) Which of the following, a 0.1 m aqueous solution will have the lowest decrease in freezing point (Tf) (Justify your answer)? The Kf for water is 1.86 °C A) K2CrO4 B) Al(NO3)3 C) Fe3(SO4)2 D) HCl E) C6H12O6 2) What is the principal reason for the extremely low solubility of NaCl in benzene (C6H6)? 1) For an electrolyte T f = i x K f x m Tf is the freezing point depression i is the van’t Hoff factor, a measure of the extent to which electrolytes dissociate in solution Kf, is the molal freezing-point-depression constant (oC/m) m is the molality of the solution Since the solutions we have to compare are all aqueous (ie solvent = water) and that their molality is identical, the comparison of their respective i is sufficient. The ideal (or limiting) values of i for each electrolyte are tabulated below: Electrolyte Fe3(SO4)2 Al(NO3)3 K2CrO4 HCl C6H12O6 van’t Hoff factor i (limiting value) 5 4 3 2 1 The lowest freezing-point-depression should be observed by the solution of electrolyte having the lowest van’t Hoff factor. C6H12O6 (sucrose) has the lowest van’t Hoff factor, it therefore has the lowest freezing-point-depression. Note: Tf (C6H12O6) < Tf (HCl) < Tf (K2CrO4 ) < Tf (Al(NO3)3 < Tf (Fe3(SO4)2) 2) The general rule is “like dissolves like”. Thus, compound having similar intermolecular forces will interact with each other and thus be soluble in one another. NaCl possesses the following intermolecular forces: ionic bonding and dispersion forces. Benzene possesses the following intermolecular forces: dispersion forces. Therefore, NaCl is polar and benzene is non-polar (non polar compounds are those that possess exclusively dispersion forces), and that is why NaCl has a poor solubility in benzene. Note: NaCl (polar compound) is very soluble in water (polar compound). Benzene (nonpolar compound) is very soluble in hexane (non-polar compound). IV When a given amount of ethylene glycol (Molar Mass = 62 g/mol) is added at 35 °C to 1.00 kg of ethanol (Molar Mass = 46 g/mol), the vapor pressure of the solution decreases to 86 torr. If the vapor pressure of ethanol at 35 °C is 100 torr, what is the mass of ethylene glycol that must be added to 1.00 kg of ethanol to reduce its pressure to 86 torr? Calculate the mole fraction, X(ETHANOL), of the solvent (ethanol): According to Raoult’s law: P o Pethanol = X ethanol Pethanol X ethanol ethanol o Pethanol 1 P X EG 1 X ethanol 1 ethanol Po ethanol n EG n EG and nTOT n EG nethanol X EG nEG P n EG 1 ethanol Po nethanol ethanol nethanol n EG 1 P 1 ethanol Po ethanol n EG nEG nethanol n EG 1 P 1 ethanol o P ethanol methanol M ethanol nethanol 1 1 1 1 Pethanol Pethanol 1 o 1 o P ethanol Pethanol Finally, n EG m EG M EG M EG 62 1 10 3 M ethanol 46 219 g m EG 1 1 1 1 86 Pethanol 1 1 100 o Pethanol methanol m EG n EG M EG Conclusion: 219 g of ethylene glycol must be added. V 1. Which of the following does not form a solution? Justify your answer? (i) NaCl(s) in liquid NH3(l) ; (ii) C12H26(s) in C3H8(l) ; (iv) H2O(l) in liquid H2S(l) (iii) CH3Cl(l) in CCl4(l); (iii) CH3Cl(l) in CCl4(l) does not form a solution because CH3Cl(l) is polar while CCl4(l) is not polar. NaCl(s) and NH3(l) can form a solution thanks to ion-dipole interactions. C12H26(s) and C3H8(l) can form a solution because they are both non-polar; they only have dispersion forces. H2O(l) in liquid H2S(l) can form a solution thanks to dipole-dipole interactions. 2. A sulfuric acid solution, H2SO4, in water has a concentration of 2.45 M. The density of water is 1.15 g/mL at 20°C. What is the molality at this temperature? (dH2SO4=1.84, MH2SO4=98.1) m V n solute in mol C V C solution solution solute solution masssolvent in kg d solvent Vsolvent d solvent Vsolvent What volume of concentrated H2SO4 is required to prepare a 1 L solution that is 2.45 M. n H 2 SO4 C H 2 SO4 Vsolution VH 2 SO4 mH 2 SO4 M H 2 SO4 d H 2 SO4 VH 2 SO4 M H 2 SO4 VH 2 SO4 C H 2 SO4 Vsolution M H 2 SO4 d H 2 SO4 2.54 1 98.1 135 mL 0.135 L Vwater 1 0.135 0.865 L 1.84 A 1 L of 2.45 M solution of H2SO4 contains 0.865 L of water, therefore: C 2.45 mol.L1 1 m solute 2.46 m d solvent 0.865 1.15 kg.L1 0.865 Conclusion: At this temperature, the molality is 2.46 m. VI Lysozyme is an enzyme that breaks bacterial cell walls. A solution containing 0.150 g of this enzyme in 210 mL of solution has an osmotic pressure of 0.953 torr at 25°C. What is the molar mass of this substance? The osmotic pressure of a solvent obeys the following law: m V nRT M Therefore, M RT mRT 0.150 0.082 (25 273) 13919 g / mol 0.953 V 3 210 10 760 Conclusion: The molar mass of this substance is 13919 g/mol.