Sept. 26, 2003                                     NAME____________________________

 

 

Chemistry 103 Section 1

EXAM #1—12:10- 1:30.

 

IMPORTANT NOTE--For partial credit be sure to show your reasoning.  Sometimes your instructor may go to heroic lengths to give you partial credit if you show work that is both legible logical.  If you need more paper, write on the back of the page and indicate clearly that you have done so.  Units and significant figures are important!

 

Look over all questions before starting and plan your time wisely. 

 

There will be no partial credit on Problem 1.

 

 

(Possibly Useful )Physical Constants

NA = 6.022 x 1023

Mass of an electron 9.1 x 10-28 g

Mass of a proton      1.67 x 10-24 g

Mass of a neutron    1.67 x 10-24 g

 

Density of water 1.00 g/cm3  

 

Solubility Table           Activity Series

 

Periodic Table

 

 

 

Problem 1________/20          Approximate Grading Scale

 

Problem 2________/25          100-90    4.0 Congratulations! Keep it up!

 

Problem 3________/25          75-89  3.0, 3.3, 3.7  Good, but practice

 

Problem 4________/30       50-74    2.0, 2.3, 2.7  Practice lots

 

 

Total        _________       20-49  1.0, 1.3, 1.7  seek help on fundamentals

 

 

 

 

Please show your work clearly and use significant figures correctly.

 

Problem 1—20 points

 

State whether the quantity on the left is greater than (>), less than ( < ), or equal (=) to the quantity on the right.  You may use either words or symbols and the first example is done for you.

 

 

The number of candidates for mayor of Philadelphia

 

      

             <

 

The number of candidates for governor of California

 

The diameter of a gold atom

 

 

         >

 

The diameter of the gold atom’s nucleus

 

 

The number of ions in 100 mL of a 1.0 M solution of NaCl

 

 

 

           <

 

The number of ions in 100 mL of a 1.0 M solution of MgCl2

 

The number of cm3 in 1 m3

 

 

          >

 

The number of cm2 in 1 m2

 

The number of electrons in a Neon atom, Ne

 

 

         =

 

The number of electrons in an oxide anion, O 2-

 

The number of moles of NaOH required to neutralize 100 mL of 0.5 M HCl

 

 

 

         <

 

The number of moles of NaOH required to neutralize 100 mL of 0.5 M H2SO4

 

 

Problem 2 (25 points)

 

Calculate the volume of ethanol, C2H6O, produced in the fermenation of 352 g of glucose, C6H12O6.  The density of ethanol is 0.789 g  / mL..

 

C6H12O6   ---->   2 C2H6O,   +  2 CO2

 

                                           180 g /mol             46 g/mol

                                     352 g x 1 mol / 180 g = 1.96 moles glucose

 

1.96 moles glucose x 2 moles ethanol / 1 mole glucose x 46 g / mol = 179.9 g ethanol

 

179.9 g ethanol x 1 mL / 0.789 g = 228 mL

 

This was a homework problem!  Use Molar masses and density as conversion factors.

 

 

Problem 3 (25 points)

 

a)              Complete the reaction below by stating what the products are:

 

10.5 g             15.5 g

208.2 g/mol    161.5 g/ mol     233 g / mol

BaCl2 (aq) + ZnSO4 (aq) à BaSO4 (s) + ZnCl2 (aq)

 

According to Solubility BaSO4chart is insoluble so this is the solid precipitate

 

A clear solution of BaCl2 is made by dissolving 10.5 g into 100.0 mL water in Beaker A and the ZnSO4 solution is made by dissolving 15.5 g of zinc sulfate in 200 mL of water in Beaker B.  The two solutions are mixed together in a big Beaker C and then solid precipitate is filtered out and dried. 

 

 

b)    How many moles of BaCl2, barium chloride are present in Beaker A before mixing?

 

 

10.5 g x 1 mol/ 208.2 g = 0.504 moles (3 sf)

 

c)   How many moles of ZnSO4 , zinc sulfate, are present in Beaker B before mixing?

 

 

15.5 g x 1 mol/ 161.5 g = 0.0960 moles (3 sf)

 

 

d)    How many moles of solid precipitate are present in Beaker C after mixing? (Hint: the molecular form of the equation above will be most helpful)

 

This is a limiting reagent problem and 0.504 moles of BaSO4 are produced.

 

e)     What is the mass of the precipitate in grams?

 

0.0504 moles x 233 g / mol = 11.7 g

 

The volumes of the solutions are irrelevant because no concentrations are calculated.

 

Problem 4 (30 points)

The average adult has 10 L of blood.

 

The HIV virus contains 2 RNA molecules surrounded by many protein molecules.  Current therapeutic strategies center on inhibiting viral replication by interfering with both HIV DNA production and the correct processing of the HIV proteins.  One drug which inhibits the latter process is called Ritonavir, C37H48N6O5S2, whose molar mass is 720.95 g/mol.

 

a).  Before treatment an infected indivual has 643,000 RNA molecules per mL of blood.  Find the molar concentration (moles virus /Liter blood) of viruses in the blood.

 

643000/ 2 x 1/mL x 1000 mL/ 1 L x 1 mol/ 6.022 x 10 ^ 23 = 5.34 x 10 ^ -16 M

 

Watch out—dividing by Avogadro’s Number should result in a very small (negative exponent ) concentration.

 

b).  One orally administered dose of ritonavir is 600.0 mg.  Assuming that 55.0% of the drug appears in the bloodstream one hour following adminstration of ritonavir, find the molar concentration of ritonavir in the blood (moles drug / Liter blood).

 

 

600 ng x .55 x 1/ 10 L x 1 g/ 1000 mg x 1 mol/ 720.95 g = 4.58 x 10 ^ -5 M

 

 

c). What is the ratio of the number of ritonavir molecules to the number of viruses?

 

4.58 x 10 ^ -5 moles drug  / 5.34 x 10 ^ -16 moles virus = 8.49 x 10 ^ 10 drug molecules per virus (in the same volume).

 

A large excess of drug is required because very few drug molecules actually reach their viral target.

 

What do you think the maximum possible concentration of anything could be?  The minimum?