If I have #1254# aluminum cans, and every #22# of them weighs #"1 lb"#, how much volume in #"L"# does this set of #1254# cans occupy?

1 Answer
Truong-Son N.
Oct 1, 2017

This problem is missing the density of aluminum, which is #"2.70 g/cm"^3#.

#V_(Al) = "9.57 L"#

Given #1254# aluminum cans and #"22 cans"/"lb"#, we have this many pounds (given that #22# is counted and thus exact):

#1254 cancel"cans" xx "1 lb"/(22 cancel"cans")#

#=# #"57.00 lbs Al"#

Knowing its density and the estimation that #"2.205 lbs"# #=# #"1 kg"#:

#57.00 cancel"lbs" cancel"Al" xx cancel"1 kg"/(2.205 cancel"lbs") xx (1000 cancel"g")/cancel"1 kg" xx "1 cm"^3/(2.70 cancel"g Al")#

#=# #"25.85 kg"# mass of aluminum cans having a #"9574 cm"^3# volume.

And so, the volume in #"L"# is more conversion:

#color(blue)(V_(Al)) = 9574.20 cancel("cm"^3) xx (cancel"1 mL")/cancel("1 cm"^3) xx "1 L"/(1000 cancel"mL")#

#=# #color(blue)("9.57 L of aluminum cans")#

limited by the sig figs in the density of aluminum.

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