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CHEM 103 – Portage Learning – Module 6 Exam Questions and Verified Answers (2024/2026) – Complete Test Preparation Material This document includes all the verified questions and answers for Module 6 of CHEM 103 from Portage Learning. It covers essential concepts related to acids and bases, pH calculations, neutralization reactions, and buffer systems. The content is aligned with the module objectives and serves as a reliable resource for achieving a high score on the exam.
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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 (C 6 H 12 O 6 ). 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 (C 6 H 12 O 6 ) 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
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. GaCl 3 , Al 2 (SO 4 ) 3 , NaI, MgCl 2 Your Answer: Al2(SO4)3 - > 2Al3+, 3SO4 -^2 = = 5 ions - > most ions = lowest freezing point GaCl2 - > Ga3+, Cl-^ = 4 ions MgCl2 = 3 ions NaI = 2 ions - > least ions = highest freezing point GaCl 3 3rd lowest FP → (^) Ga+3 + 3 Cl-^ ∆tf = 1.86 x 0.1 x 4 = Al 2 (SO 4 ) 3 lowest FP → 2 Al+3^ +
exam.
Question 8
Show the calculation of the mass of Ba(NO 3 ) 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. 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(NO 3 ) 2. molessolute = gsolute / MW molessolute = 37.5 g / 261.55 = 0.1434 mol Molarity = moles / (mL /1000)
Question 9
Question 10
mL / 1000 = 0.1434 / 0.667 = 0. 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 C 4 H 8 O 4 in 250 grams of water. Kb for water is (^) 0.51, BP of pure water is 100oC. Calculate your answer to 0.01oC. molality = (gsolute / MW) / (gsolvent / 1000) molality = (20.9 / 120.104) / (250 / 1000) = 0.6961 m ∆tb = Kb x m = 0.51 x 0.6961 = 0.355oC BPsolution = BPsolvent - ∆tb = 100 oC + 0.355 = 100.35oC 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^ - 1.35oC. 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. 0.1452 = (14.5 / MW) MW = 14.5 / 0.1452 = 99.