BELLARMINE COLLEGE
DEPARTMENT OF CHEMISTRY AND PHYSICS
CHEM115 - GENERAL, ORGANIC, AND BIOLOGICAL CHEMISTRY II
EXAM #2
3/ 10/00
ANSWER KEY
Do all the following questions. Please read each question carefully and apportion your time according to the relative
worth of the question (shown in parenthesis after the question). Write legibly. Marks will be deducted for unclear,
messy and illegible work. You must show all calculations (including units) where appropriate. Also, remember to
show answers to the correct number of significant figures. Answer in the space provided after each question. The
total marks for this exam is 50.
1.
A 100.0-mL bubble of hot gases at 225.0oC and 1.85 atm escapes from an active volcano. What is the new
volume of the bubble outside the volcano where the temperature is -25.5oC and the pressure is 1.05 atm? (5)
PiVi/Ti = PfVf'/Tf and Vf = PiViTf/TiPf = (1.85 atm x 100.0 mL x 247.5 K)/(498 K x 1.05 atm)
= 87.6 mL (to 3 SF)
2.
Assume you have 1.75 g of the deadly gas HCN, hydrogen cyanide. What is the volume of the gas at STP?
(2.5)
1.75 g HCN x (mol HCN/27.03 g HCN) x (22.4 L HCN/mol HCN) = 1.45 L (to 3 SF)
3.
a.
b.
Gas pressure outside the space shuttle is approximately 1.0 x 10 -14 mm Hg at a temperature of around 1.0K. (7)
If the gas consists of almost only hydrogen atoms (H, not H 2) what volume of space is occupied by 1.0 mol of
atoms? (1.0 atm = 760.0 mm Hg, and R = 0.0821 L.atm/mol.K)
What is the density of H gas in atoms per liter?
a.
1.0 x 10-14 mm Hg x (1.0 atm/760.0 mm Hg) = 1.32 x 10-17 atm
PV = nRT and V = nRT/P = (1.0 mol x 0.0821 atm.L/mol.K x 1.0 K)/ 1.32 x 10-17 atm
= 6.2 x 1015 L (to 2 SF)
b.
(1.0 mol/6.2 x 1015 L) x (6.02 x 1023 atoms/mol) = 6.02 x 1023 atoms/6.2 x 1015 L = 9.7 x 107 atoms/L (to 2 SF)
4.
Assuming a total pressure of 9.5 atm, what is the partial pressure of each component in the mixture of 98%
helium and 2.0% oxygen breathed by deep-sea divers? (2.5)
pHe = 98/100 x 9.5 atm = 9.3 atm pO2 = 2/100 x 9.5 atm = 0.2 atm
5.
A patient receives all of her nutrition from fluids given intravenously. Every 12 hours, 800.0 mL of a solution
that is 4.0% (m/v) amino acids from proteins and 26.0% (m/v) glucose is given along with 450.0 mL of a 10.0%
(m/v) lipid (fat). In 1 day, how many grams of amino acids, glucose and lipids are given to the patient? (7)
Amino acid: (800.0 mL/12 hrs) x (4.0 g/100 mL) x (24 hrs/12 hrs) = 64 g/24 hrs
Glucose: (800.0 mL/12 hrs) x (26.0 g/100 mL) x (24 hrs/12 hrs) = 416 g/24 hrs
Lipid: (450.0 mL/12 hrs) x (10.0 g/100 mL) x (24 hrs/12 hrs) = 90 g/24 hrs.
Total = 570 g/day
6.
What is an electrolyte?
Give a couple of examples. (3)
Any substance that when dissolved in water permits the flow of electricity through the resulting solution.
NaCl and acetic acid are two examples of electrolytes.
7.
Why does a red blood cell swell up and burst when placed in pure water? (3)
Pure water contains no solutes dissolved in it whereas the cytoplasm of the red blood cell does contain a
variety of solutes dissolved in it. The pure water will therefore enter the red blood cell through the cell's
membrane (a semi-permeable membrane) in order to reduce the concentration of solutes in the
cytoplasm. This happens because the most stable situation is when both the internal and external
concentrations are the same (no concentration gradient). The influx of water into the red blood cell
causes the cell to swell and eventually burst.
8.
Which of the following solutions is more concentrated? 2.5% (m/v) NaHSO 4 or 0.025 M NaHSO4. (4)
Need to express one concentration in terms of the other or both in a third concentration so a
comparison can be made.
(0.025 mol NaHSO4/L) x (120.05 g NaHSO4/mol) = 3.00 g NaHSO4/L => 0.30 g NaHSO4/100 mL
= 0.30 % (m/v)
Therefore, 2.5 % (m/v) NaHSO4 is more concentrated that 0.025 M NaHSO4 [0.30% (m/v)]
9.
One test for vitamin C (C6H8O6, also called ascorbic acid) is based on the oxidation of the vitamin by iodine:
C6H8O6(aq) + I2(aq)  C6H6O6(aq) + 2HI(aq)
a.
If 25.0 mL of a fruit juice requires 13.0 mL of 0.0100 M I 2 solution for reaction, what is the molarity of the
ascorbic acid in the fruit juice?
Moles C6H8O6 = moles I2 (from the stoichiometry of the balanced chemical reaction.
Moles I2 = molarity I2 (mol/L) x volume (L) = 0.0100 mol I2/L x (13.0 ml x (L/1000 mL)) = 1.3 x 10 -4 mol I2
1.3 x 10-4 mol I2 x (mol C6H8O6/mol I2) = 1.3 x 10-4 mol C6H8O6
Molarity C6H8O6 = 1.3 x 10-4 mol C6H8O6/(25.0 mL x (L/1000 mL)) = 5.20 x 10-3 M to (3 SF)
This simplifies to M1V1 = M2V2 where M1 and V1 are the molarity and volume, respectively, of one of the
reactants (i.e., the vitamin C) and M 2 and V2 are the molarity and volume of the second reactant (I2).
Since we want to find M1 we can rearrange the equation to solve for M 1, M1 = M2V2/V1
M1 = (13.0 mL x 0.0100 mol/L)/25.0 mL = 5.20 x 10 -3 M
b.
The FDA recommends that 60.0 mg of vitamin C be consumed per day. How many milliliters of the fruit juice
in part a must a person drink to obtain the recommended dosage? (7)
(5.20 x 10-3 mol/L) x (176.13 g C6H8O6/mol C6H8O6) x (1000 mg/g) x (L/1000 mL)
= 915.9 mg C6H8O6/1000 mL
(1000 mL/915.9 mg C6H8O6) x 60.0 mg C6H8O6 = 65.5 mL (to 3 SF)
10. In many states (though not Kentucky), a person with a blood alcohol concentration of 0.080% (v/v) is
considered legally drunk. (9)
a. What volume of total alcohol does this concentration represent, assuming a blood volume of 5.0 L?
5.0 L blood x (1000 mL/L) x (0.080 mL alcohol/100 mL blood) = 4.0 mL alcohol
b.
How much wine (in mL) that contains 13.5% (v/v) of alcohol would contain the amount of alcohol calculated in
part a?
4.0 mL alcohol x (100 mL wine/13.6 mL alcohol) = 29.6 mL wine
c.
In Kentucky, a person with a blood alcohol level of 0.10% (v/v) is considered to be legally drunk. How much
more of the wine in part b would be needed to get to this volume of alcohol?
5.0 L blood x (1000 mL/L) x (0.10 mL alcohol/100 mL blood) = 5.0 mL alcohol
5.0 mL alcohol x (100 mL wine/13.5 mL alcohol) = 37.0 mL wine
37.0 mL - 29.6 mL = 7.4 mL more wine