Chemistry 1201 Section 4
Professor Russo
Spring 1999
Exam #11
Type of exam: In class. You can use your cheat sheet and a calculator. In most cases,
vital constants and conversion factors are supplied. If you cannot see the periodic table
from where you are seated, ask one of the proctors to provide you a periodic table.
Covers: Chapter 3, 4 & 13. Problems are modifications of the take-home exams or
other assigned problems.
Duration: 45 minutes.
Value: 70 points. Each question is 7.77 points. Remember….you are trying to earn 700
points on all in-class and take-home exams. After this, we will have had 1000 total
possible points.
Instructions: Mark your answers on a LARGE scantron using a #2 pencil. Be sure to
write and bubble in your name and student ID number! Fill out the scantron just
before turning it in. You should mark the right answers directly on this exam until you
are sure you have the answer you wish to select.
Correct Answers: Will show immediately after exam. So stick around if you finish
early!
________________________________________________________________________
Question #1. (BLB 3.103) Calculate the amount of KCN (potassium cyanide) required
to produce a lethal dose of HCN (300 mg HCN/kg of air) when reacted with excess acid
in a room having a volume of 6.4 x 104 liters. Assume T= 26oC where the density of air
is 0.00118 g/cm3.
a) 9.4 g KCN
b) 18.8 g KCN
c) 54.5 g KCN
d) 1090 g KCN
e) 10.9 g KCN
Chem 1201 Sec. 4
Exam 11
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Question #2. (BLB 10.26) The Goodyear Blimp Eagle holds 202,700 cubic feet of
Helium gas, which cannot explode (The Eagle is about four times smaller than the
Hindenburg, carries much less cargo and it is slower---ah, progress!). Compute the mass
of Helium gas in grams, assuming 76oF and 1 atm pressure.
a) 14,000,000 g
b) 3,700,000 g
c) 1,400,000 g
d) 940,000 g
e) 235,000 g
Question #3. (based on BLB 10.42) Ethanol, C2H6O, "burns" in our bodies using
oxygen to produce CO2 and H2O. Balance the reaction and calculate the volume of CO2
produced at body temperature of 37oC and pressure of 1.00 atm when 5.00 g ethanol is
metabolized. You can assume all water vapor is removed from the CO2 prior to
measurement (i.e., we want the "dry volume" of CO2).
a)
b)
c)
d)
e)
5.5 liters CO2
2.75 liters CO2
0.660 liters CO2
558 liters CO2
1.05 liters CO2
Chem 1201 Sec. 4
Exam 11
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Question #4. (similar to BLB 10.92 or 10.91) A particular gas is composed of 85.7 % C
and 14.3 % H by mass. If 3.42 g of the gas has a volume of 2.00 liter at 615 torr and
50.0oC, what is the molecular formula of the gas?
a) C2H4
b) CH2
c) C2H6
d) C4H8
e) C8H16
Question #5. (like BLB 4.8b) Compute the molarity of a solution that contains 0.933 g
Na3PO4 in enough water to form 500.0 mL of total solution. The molarity of the sodium
ions is:
a) 1.035 M Na+
b) 0.0114 M Na+
c) 0.00557 M Na+
d) 0.256 M Na+
e) 0.0341 M Na+
Chem 1201 Sec. 4
Exam 11
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Question 6. (based on BLB 4.8) The number of grams of nitrogen present in 38.0 mL
of 3.05 M Ba(NO3)2 is:
a) 1.87 g N
b) 3.24 g N
c) 4.28 g N
d) 1.62 g N
e) 5.84 g N
Problem 7 (BLB 4.52) In a titration, how many mL of 0.155 M H2SO4 are needed to
neutralize completely 68.0 mL of 0.101 M Ba(OH)2 ?
a) 22.2 mL
b) 104.3 mL
c) 53.8 mL
d) 88.6 mL
e) 44.3 mL
Chem 1201 Sec. 4
Exam 11
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Problem 8 (Based on BLB 13.56) Assuming full dissociation of the following
compounds, which has the lower freezing point?
a) 1.0 molal solution of H2SO4
b) 1.0 molal solution of HCl
c) 0.5 molal solution of NaNO3
d) 0.25 molal solution of Ca(NO3)2
e) 0.25 molal solution of FeCl3
Problem 9 (Based on Ch 4-13 Notes) Here is a problem you may not have seen yet
unless you read the latest Ch. 4-13 notes carefully. But relax--it is really just an osmotic
version of the ideal gas problem for molecular weight…and that's a good one for you to
know! To solve the problem, remember V = nRT where  = osmotic pressure.
Lysozyme is an enzyme (protein) that breaks bacterial cell walls. A 210-mL solution
containing 0.150 g of lysozyme has an osmotic pressure of 0.953 torr at 25oC. What is
the molecular weight of lysozyme?
a)
b)
c)
d)
e)
1.83 g/mol
1170 g/mol
13,900 g/mol
9,400 g/mol
340 g/mol
Chem 1201 Sec. 4
Exam 11
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