CHEM 103 MODULE 6 2022

Question 1 Not yet graded / 10 pts Question 2 Not yet graded / 10 pts Question 3 Not yet graded / 10 pts CHEM 103 MODULE 6 2022 Click this link to access the Periodic Table. This may be helpful throughout the exam. List and explain if each of the following solutions conducts an electric current: sodium chloride (NaCl), hydrochloric acid (HCl) and sugar (C6H12O6). Sodium chloride (NaCl) is an ionic compound and conducts since it forms ions in solution. Hydrochloric acid (HCl) is a polar compound and conducts since it forms ions in solution. Sugar (C6H12O6) is a molecular compound but does not form ions in solution so it does not conduct. Explain how and why the presence of a solute affects the boiling point of a solvent. The presence of a solute raises the boiling point of a solvent by lowering the vapor pressure of the solvent. With this lower vapor pressure, more heat (a higher boiling point) is required to raise the vapor pressure to atmospheric pressure. Click this link to access the Periodic Table. This may be helpful throughout the exam. Question 4 Not yet graded / 10 pts Rank and explain how the freezing point of 0.100 m solutions of the following ionic electrolytes compare, List from lowest freezing point to highest freezing point. GaCl3, Al2(SO4)3, NaI, MgCl2 Your Answer: Al2(SO4)3 -> 2Al 3+ , 3SO4 -2 = = 5 ions -> most ions = lowest freezing point GaCl2 -> Ga 3+ , Cl - = 4 ions MgCl2 = 3 ions NaI = 2 ions - > least ions = highest freezing point GaCl3 3rd lowest FP → Ga+3 + 3 Cl - ∆tf = 1.86 x 0.1 x 4 = Al2(SO4)3 lowest FP → 2 Al +3 + 3 SO4 -3 ∆tf = 1.86 x 0.1 x 5 = NaI = highest FP → Na + + I - ∆tf = 1.86 x 0.1 x 2 MgCl2 = 2nd lowest FP → Mg+2 + 2 Cl - ∆tf = 1.86 x 0.1 x 3 FP: Al2(SO4)3 < GaCl3 < MgCl2 < NaI Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the mass percent solute in a solution of 20.8 grams of Ba(NO3)2 in 400 grams of water. Report your answer to 3 significant figures. Question 5 Not yet graded / 10 pts Question 6 Not yet graded / 10 pts Question 7 Not yet graded / 10 pts Mass % = (20.8 / 20.8 + 400) x 100 = 4.94% Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the molality of a solution made by dissolving 28.5 grams of C8H16O8 in 400 grams of water. Report your answer to 3 significant figures. molality = (gsolute / MW) / (gsolvent / 1000) molality = (28.5 / 240.208) / (400 / 1000) = 0.297 m Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the molarity of a solution made by dissolving 35.9 grams of Mg(NO3)2 to make 400 ml of solution. Report your answer to 3 significant figures. Molarity = (gsolute / MW) / (mlsolvent / 1000) Molarity = (35.9 / 148.325) / (400 / 1000) = 0.605 M Click this link to access the Periodic Table. This may be helpful throughout the exam. Question 8 Not yet graded / 10 pts Show the calculation of the mass of Ba(NO3)2 needed to make 250 ml of a 0.200 M solution. Report your answer to 3 significant figures. Molarity = (moles) / (mlsolvent / 1000) 0.200 = (moles) / (250 / 1000) Moles = 0.200 x 0.250 = 0.0500 Moles = (gsolute / MW) 0.0500 = (gsolute / 261.55) gsolute = 0.0500 x 261.55 = 13.1 g Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the volume of 0.667 M solution which can be prepared using 37.5 grams of Ba(NO3)2. molessolute = gsolute / MW molessolute = 37.5 g / 261.55 = 0.1434 mol Molarity = moles / (mL /1000) 0.667 = 0.1434 / (mL / 1000) Question 9 Not yet graded / 10 pts Question 10 Not yet graded / 10 pts mL / 1000 = 0.1434 / 0.667 = 0.2150 mL = 0.2150 x 1000 = 215 mL Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the boiling point of a solution made by dissolving 20.9 grams of the nonelectrolyte C4H8O4 in 250 grams of water. Kb for water is 0.51, BP of pure water is 100 oC. Calculate your answer to 0.01 oC. molality = (gsolute / MW) / (gsolvent / 1000) molality = (20.9 / 120.104) / (250 / 1000) = 0.6961 m ∆tb = Kb x m = 0.51 x 0.6961 = 0.355 oC BPsolution = BPsolvent - ∆tb = 100 oC + 0.355 = 100.35 oC Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the molar mass (molecular weight) of a solute if a solution of 14.5 grams of the solute in 200 grams of water has a freezing point of -1.35 oC. Kf for water is 1.86 and the freezing point of pure water is 0 oC. Calculate your answer to 0.1 g/mole. ∆tf = Kf x m molality = ∆tf / Kf = 1.35 / 1.86 = 0.726 m molality = (gsolute / MW) / (gsolvent / 1000) 0.726 = (moles) / (200 / 1000) Moles = 0.726 x 0.200 = 0.1452 0.1452 = (14.5 / MW) MW = 14.5 / 0.1452 = 99.9