Question

The aluminum foil on a certain roll has a total area of 18.5 m squared and a mass of 1275 g. Using density of 2.7 g per cubic centimeter for aluminum, determine the thickness in millimeters of the aluminum foil.

Answer 1

Answer is: the thickness is 0.0256 millimeters.

m(Al) = 1275 g; mass of aluminium foil.

d(Al) = 2.7 g/cm³; density of aluminium foil.

V(Al) = m(Al) ÷ d(Al).

V(Al) = 1275 g ÷ 2.7 g/cm³.

V(Al) = 472.2 cm³ ÷ 10⁶ cm³/m³.

V(Al) = 0.000472 m³, volume of aluminum foil.

V(Al) = P · l (thickness).

l = 0.000472 m³ ÷ 18.5 m².

l = 0.0000256 m · 1000 mm/m.

l = 0.0256 mm.