 # The stearic acid/hexane solution

Each 1 mL of stearic acid/hexane solution contains 0.1 g of stearic acid. What is this solution’s molarity?
Molarity is a measure of a solvent’s concentration which is proportional to moles of the solute being dissolved separated by liters of the solution in which the solute is dissolved (Timberlake & Frost, 2016).
Mole of stearic acid=1g is 1.86*10-7mol
0.1g = ?
1.86*10-7mol * 0.1g
=1.86*10-8moles
0.1g stearic acid=1.86*10-8moles
Hexane solution=0.001 liters of solution
molarity=1.86*10-8moles/0.001 liters
molarity=1.86*10-5 mol/liters
What is the largest area of a circle that can be made from a piece of thread 25.2 cm long (assume no knot)?
25.2cm=2π r 25.2cm=2×3.14×r
r=4.013cm
Area = π r2
=3.14×4.013×4.013
ans=50.56cm2

1. What is volume of one aluminum atom? (show your work)

Mass of aluminum=0.89g
Density of aluminum=2.70 g/cm3
Volume=Mass/Density
=0.89/2.70
Volume=0.330cm3

2. Research face-centered cubic structures on the web. What is the length of one side of the crystal unit cell for aluminum? (show your work)
Edge length=a
a=2r
2r=2(143×10-12)
Ans =2.86×10-10

Laboratory Questions
Stearic Acid
1. Describe the appearance of the monolayer and the thread, as the first drops of stearic acid were added to the water.
The polar end (acid end of the molecule) dissolves in water while the non-polar portion (hydrocarbon portion of the molecule) tends to stick out of the water.

2. Describe the appearance of the monolayer and thread as the last drops of stearic acid were added to the water.
The molecules are arranged in a single layer vertically as close as possible.

3. If some the hexane had evaporated from the stearic acid solution, what effect (if any) would it have on the experimental value of Avogadro’s number? Why?
It would have affected Avogadro’s number as it would have altered the concentration of stearic acid in hexane thus altering Avogadro’s number.

4. What assumptions are made in the stearic acid experiment to calculate the value of Avogadro’s number?

• time taken for stearic acid to evaporate
• use of different solvent

5. What is the percent error for your experimental determination of Avogadro’s number?

Percent Error: 99.99%

6. What are three possible sources of error in the stearic acid experiment?

• Evaporation of stearic acid too drastic
• Volume of drops used per trial
• Use of different solvent for dilution

Aluminum Foil
7. Write out a step-by-step procedure for determining Avogadro’s number using the provided data.
• You get the length and width of the aluminum foil so as to determine the area of the aluminum foil.
• You then weigh the aluminum foil to get the mass.
• You divide the area of the aluminum foil with the area of one stearic acid molecule to determine the number of aluminum atoms.
• You finally divide the number of aluminum atoms with the moles of aluminum to get Avogadro’s number

8. Based on your data, what value for Avogadro’s number was calculated?
Cm/moles

9. What is the percent error for your experimental determination of Avogadro’s number?

74%

10. Include all data and calculations to determine the layers of atoms, assuming each aluminum atom is stacked on top of one another and are touching.

Mole of aluminium=0.89/26.98
=0.033mol
58.34cm2/20.7 × 10-16 cm2

=2.8184×1016

11. Include all data and calculations to determine the layers of atoms, assuming a face-centered cubic (FCC) structure.

Edge length=a
a=2r
2r=2(143×10-12)
Ans =2.86×10-10

12. What are some possible sources of error in the aluminum foil experiment?

• Faulty measurements
• Miscalculations

Determination of Avogadro’s Number Using Aluminum Foil

Length
(cm) Width
(cm) Area
(cm2) Mass
(g)
18.82cm 3.1cm 58.34cm2 0.89g

1. Determine the volume of the aluminum foil using the density of foil.

Mass of aluminum=0.89g
Density of aluminum=2.70 g/cm3
Volume=Mass/Density
=0.89/2.70
Volume=0.330cm3

2. Determine the volume of one atom of aluminum.

26.98 g/mol

3. Determine the number of atoms of aluminum in the foil sample.

58.34cm2/20.7 × 10-16 cm2

=2.8184×1016

4. Determine the moles of aluminum.

Mole of aluminium=0.89/26.98
=0.033mol

Mole of aluminium=0.89/26.98
=0.033mol
=8.541×1017

Layers Calculation
1. Determine the thickness of the aluminum foil.

2.70 g/cm3

2. If aluminum atoms are stacked on top of each other and one aluminum atom has a diameter of 286 pm or 2.86 × 10-8 cm, and calculate how many layers are in the foil.

Mole of aluminium=0.89/26.98
=0.033mol
58.34cm2/20.7 × 10-16 cm2

=2.8184×1016

3. If the unit cell is 406 pm or 4.06 × 10-8 cm and the height is 1.48 × 10-3 cm:

794cm2/20.7 × 10-16 cm2
=3.8357×1017

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