The problem that follows was given by an instructor to his introductory chemistr
ID: 986692 • Letter: T
Question
The problem that follows was given by an instructor to his introductory chemistry class. He asked them to calculate the cost of a single aluminum atom in a roll of aluminum foil that he had recently purchased. The information that he provided the class was the following: 1roll of Reynonlds Wrap Heavy Duty Aluminum 18.0 in wide, 450.0 ft long and 0.90mil thick The cost $45.98
That is all the information he gave the class, now they were on their own. Two students, became quite upset with the problem they were assigned and withdrew from the course, several students went to complain to the instructor, and others to the Department Chair. They all insisted that the problem was impossible to solve, but you know better. Describe how the solve the problem.
Explanation / Answer
Volume of the aluminium atom = (Radius of atom)^3 = (115 pm)^3 = 1.520875 * 10^(-30) m^3
Volume of the sheet = 18 inch * 450 ft * 0.90 mil
=> 0.4572 m * 137.16 m * 0.0009 m
=> 0.05643 m^3
Number of aluminium atoms = 0.0564 m^3/1.520875 * 10^(-30) m^3 = 3.7109 * 10^(28) atoms
Mass of aluminium atom = 45.98/(3.7109 * 10^(28)) = 1.239 * 10^(-27) $
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