Sample%20Exam%20III%20Solutions

Sample%20Exam%20III%20Solutions - Department of Chemistry...

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Department of Chemistry CHM 1220/1225 Exam III – Sample Exam Solutions 1. T 1 = 28.0 + 273.15 = 301.2 K P 1 = 400.0 mmHg T 2 = 46.5 + 273.15 = 319.7 K P 2 = 760.0 mmHg (at the normal boiling point) Δ = K . K . K mol J . H mmHg . mmHg . ln vap 7 319 1 2 301 1 3145 8 0 400 0 760 = Δ mmHg . mmHg . ln K . K . K . K . K mol J . H vap 0 400 0 760 2 301 7 319 7 319 2 301 3145 8 Δ H vap = 2.778 x 10 4 J/mol or 27.78 kJ/mol 2. Consider the three molecules: H 2 S, F 2 , and CH 3 OH. a. Identify the intermolecular forces present for each: H 2 S LONDON FORCES, DIPOLE-DIPOLE FORCES F 2 LONDON FORCES CH 3 OH LONDON FORCES, DIPOLE-DIPOLE FORCES, HYDROGEN BONDING b. Arrange the molecules from weakest to strongest intermolecular forces. F 2 < H 2 S < CH 3 OH c. Which has the highest vapor pressure? F 2 d. Which has the highest boiling point? CH 3 OH e. Which has the highest heat of vaporization? CH
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Sample%20Exam%20III%20Solutions - Department of Chemistry...

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