How Much Does a Cubic Foot of Gold Weigh?

Answer

To know how much a cubic foot of gold does weigh will depend on the actual density of the gold. The density is what is used to determine the karat of the gold. For a cubic foot of 24 karat gold the weight would come in at 1203.74 pounds. This is with a density of 19.282 g/cm3. The weight will vary of course based on the karat as well as the overall purity. However in its purest form this is what it will weigh.
Q&A Related to "How Much Does a Cubic Foot of Gold Weigh?"
Pure gold has a density of 19.32 grams per cubic centimeter. A cubic foot contains 28,316.85 cubic centimeters. Thus, a cubic foot of pure gold has a mass of 547,081 grams, which
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I Cubic Foot of Gold would weigh 548.18 Kilograms or 1206 pounds. That is
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1 cubic foot = 0.0283 m³. density gold 19300 kg/m³. 0.0283 m³ x 19300 kg/m³ = 546 kg or 1205 US lb.
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One cubic foot of gold weighs about 1204.8586 pounds (546.51424 kg) Or about 1/3rd of an average car! report this answer. Updated on Wednesday, February 01 2012 at 09:07AM EST. Source
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