Chemistry 1A, Fall 2012 Midterm Exam #1 September 19, 2012 (90 min, closed book) Name:__________________________________ SID:___________________________________ GSI Name:________________ Discussion Day/Time: ________________ • The test consists of 4 short answer questions and 17 multiple choice questions. • Put your written answers in the boxes provided. Answers outside the boxes may not be considered in grading. • Write your name on every page of the exam. Question Page Points Score Multiple Choice (1-17) 2-6 51 Ethylene Glycol Short Answer 3 4 Boron Nitride Short Answer 4 4 Allene Short Answer 5 4 Cyanogen Short Answer 5 7 Total 70 Useful Equations and Constants: PV = nRT Ptotal = PA + PB = XAPA° + XBPB° N0 = 6.02214 x 1023 mol-1 T (K) = T (°C) + 273.15 R = 0.0821 L atm K-1 mol-1 1 atm = 760 mm Hg = 760 torr ≈ 1 barName:_____________________GSI:__________ Page 2 of 6 INTERMOLECULAR ATTRACTIONS Substance Molecular formula Structural formula Molar mass (g/mol) Boiling point ethylene glycol HOCH2CH2OH 62 197ºC propanol CH3(CH2)2OH 60 98ºC propylamine CH3(CH2)2NH2 59 48ºC trimethylamine (CH3)3N 59 3ºC 1. Which statement provides the most compelling explanation of the data? A) O−HO is stronger than H−NH B) O−HO is weaker than H−NH C) Boiling point increases with molar mass. D) Increasing the number of CH3 groups decreases the boiling point. 2. Which explanation(s) can be used to account for the lower boiling point of trimethylamine compared with the other molecules in the table? A) Trimethylamine is not polar. B) Trimethylamine does not form hydrogen bonds. C) The London dispersion forces are relatively small. D) Both B and C are correct. E) A, B, and C are all correct. 3. How would you expect the van der Waals constants to compare for propanol and propylamine? A) The constant “a” is larger for propanol, but “b” is about the same for both. B) The constant “a” is smaller for propanol, but “b” is about the same for both. C) Both constants are larger for propanol. D) Both constants are smaller for propanol. OHCCHHHHOHOHCCHHHHCHHHN HCCHHHHCHHHHNCC CHHHHHH HHHName:_____________________GSI:__________ Page 3 of 6 4. Which is the best estimate for the boiling point of ethylenediamine, NH2CH2CH2NH2? A) 35°C B) 118°C C) 192°C D) 247°C SHORT ANSWER: The boiling point of ethylene glycol is much higher than the boiling point of propanol. Draw a structure showing the interaction between ethylene glycol molecules to explain its high boiling point. BORON COMPOUNDS Substance Formula Structural formula Molar mass Boiling point Solubility in water B-O bond length boric acid B(OH)3 62 300°C 1.0 M 1.36Å Boron-oxygen bond lengths B−O 1.36 Å B=O 1.30 Å BΞO 1.20 Å 5. Which model provides the best description of the bonding in B(OH)3 based on the information provided? A) The solubility in water indicates an ionic solid with B+ and OH−. B) The high boiling point indicates network covalent bonding. C) The bonding is molecular covalent. The boron has an empty p orbital. D) The bonding is molecular covalent. The boron-oxygen bonds have a bond order of 1⅓, and the bonding satisfies the octet rule.Name:_____________________GSI:__________ Page 4 of 6 6. When B(OH)3 is added to water, a reaction occurs and B(OH)4− forms. Which statement describes the properties of the solution? A) The solution remains neutral because the concentration of H+ is equal to the concentration of OH−. B) The solution is acidic because there is an extra H+ from H2O. C) The solution is basic due to the OH− on B(OH)4−. D) B(OH)3 is not an electrolyte; B(OH)3 is a nonpolar molecule. 7. Boron trioxide, B2O3, boils at 1860°C. The compound is soluble in water. Which statement below is a correct conclusion that you can draw from the data? A) B2O3 is a relatively hard solid, but it fractures readily. B) The bonding in the solid is not very strong. C) The atoms are linked together as molecules that satisfy the octet rule. D) B2O3 is soluble in water because B3+ is so small, and the B3+ and O2- attractions in the solid are relatively weak. 8. Which bond is the most polar? A) C−N B) B−N C) B−C D) B−O SHORT ANSWER: A sketch of the structure of a layer of boron nitride, BN, is shown to the right. The structure is very similar to that of the graphite form of carbon with a B−N bond order of 1.5. However, BN is harder than graphite. Explain why. HTTP://WWW.DOCBROWN.INFO/PAGE04/4_72BOND/BNHEXAGONAL2.GIFName:_____________________GSI:__________ Page 5 of 6 ALLENE, C3H4, AND CYANOGEN, C2N2 The structure of the allene molecule is shown to the right. 9. The hybridization around the carbon atom in the middle of the molecule is A) sp B) sp2 C) sp3 D) sp4 10. The H-C=C bond angle is approximately A) 90° B) 104.5° C) 109° D) 120° 11. Which molecule is polar? A) C2H2 B) C3H4 (allene) C) N2H2 D) CF4 SHORT ANSWER: Explain why allene, C3H4, is not planar. SHORT ANSWER: Cyanogen is a toxic compound with a molecular formula C2N2. Draw two Lewis dot structures for cyanogen that satisfy the octet rule. Assign formal charges to all atoms. Circle the structure that is more stable. Structure 1 Structure 2 C C CHHHHName:_____________________GSI:__________ Page 6 of 6 VAPOR PRESSURE Equal volumes of pure water, pure ethanol, and two aqueous solutions were each injected into four identical, sealed, evacuated containers. The equilibrium vapor pressure data are given in the table below. Liquid H2O (l) CH3OH (l) CH3OH (aq) NaCl (aq) Vapor pressure at 20°C 0.023 atm 0.128 atm 0.035 atm 0.017 atm 12. What is the composition of the vapor above NaCl (aq)? A) H2O B) NaCl C) H2O and NaCl 13. What is the composition of the vapor above CH3OH (aq)? A) H2O B) CH3OH C) H2O and CH3OH 14. The rate of condensation of gaseous water is the greatest over which liquid? A) H2O (l) B) NaCl (aq) C) CH3OH (aq) D) The rates are equal for all 3 TAKE A BREATH Conditions at sea level: 1 atm and 298 K Conditions near the summit of Mount Everest: 0.4 atm and 253 K 15. Imagine that you fill your lungs to a volume of 0.5 L with air at sea level. The air is 20% O2. How many moles of O2 are in one breath? A) 0.004 moles B) 0.008 moles C) 0.02 moles D) 0.04 moles 16. Imagine that you fill your lungs to a volume of 0.5 L with air near the summit