If the average number of drops/ml = 73.0 and the overall average diameter of ole
ID: 850071 • Letter: I
Question
If the average number of drops/ml = 73.0 and the overall average diameter of oleic acid film = 38.0 cm, what would be the calculations for the following:
a. average volume of oleic acid in film
b. volume of olei acid per drop of solution
c. average area of oleic acid film
d. height of oleic acid film
e. volume of one oleic acid molecule
f. Avogadro's number
g. percent disagreement
Acceptable experimental values can be considered as within one power of ten from Avogadro's number. With this in mind what are the smallest and largest experimental results you can obtain. What are the smallest and largest percent disagreements you can obtain. I have the first half done up to g but I don't know how to figure out these last two questions.
Explanation / Answer
Given
average number of drops/mL = 73.0
Average volume of one drop of oleic acid solution:
V1 = 1.00 mL / average # of drops
= 1.00 mL / 73.0
= 0.014 mL
Average volume of oleic acid in film in cm^3
(V2) = V1 x (0.0050 mL / 1 mL) x (1 cm^3 / 1 mL)
= (0.014 mL)(0.005 mL/1 mL)(1 cm^3/1mL)
= 6.85*10^-5 cm^3
Height of oleic film in cm:
H = V2 in cm^3 / Average area of film in cm^3*1 mL = 1 cm^3
=(6.85*10^-5 cm^3/ 4.89*10^-3 cm^3)*mL
= 1.4 x 10^-2 mL
Volume of one oleic acid molecule in cm^3. A cubic shape is assumed:
V3 = H^3
= (1.4 x 10^-2 mL)^3
= 2.74*10^-6 mL^3
Avogadro's number: N = (282 g/mol) / (V3 x 0.895 g/cm^3)
= (282 g/mol) / (2.74*10^-6 mL^3)(0.895 g/cm^3)
= 1.43 x 10^24
Percentage disagreement:
% dsgmt = [log(6.0225 x 10^23) - log(N)] / log(6.0224 x 10^23) x 100
= [log(6.0225 x 10^23) - log(1.43*10^24)] / log(6.0224 x 10^23) x 100
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