The problem that follows was given by an instructor to his introductory chemistr

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 foil 24.0 in wide, 500.0 ft long and 0.96 mil thick The cost $59.48 That is all the information the instructor gave the class. 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 you would solve the problem.

Explanation / Answer

Aluminum foil is 24.0 in wide, 500.0 ft long and 0.96 mil = 0.96 * 10^-3 inch thick

Hence, the volume of the aluminium foil = widht * length * thickness

= 24.0 in * (500.0 * 12 in) *  0.96 * 10^-3 in = 138.24 inch^3

The density of aluminium = 0.098 lb/inch^3

hence, the mass of aluminum foil = 138.24 inch^3 * 0.098 lb/inch^3 = 13.547 lb

1 lb = 453.592 g

So, 13.547 lb = 13.547 * 453.592 g = 6144.81 g

Atomic weight of aluminium is 26.98 g/mol

Hence, in the aluminum foil number of moles of Al = 6144.81 g/26.98 g/mol = 227.75 mol

Now, we know 1 mol of substance contains 6.023 * 10^23 number of atoms

i.e. the aluminium foil contains 227.75 * 6.023 * 10^23 number of atoms

So, 227.75 * 6.023 * 10^23 numner of Al atoms cost $59.48

Hence, a single Al atom will cost $59.48/(227.75 * 6.023 * 10^23) = $4.34 * 10^-25

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