Problem: The average concentration of gold in seawater is 100 fM (femtomolar).  Given that the price of gold is $1764.20 per troy ounce (1 troy ounce = 31.103 g), how many liters of seawater would you need to process to collect $7000 worth of gold, assuming your processing technique captures only 50 % of the gold present in the samples? Express your answer using one significant figure.

FREE Expert Solution

amount of gold = 100 fM (femtomolar)
price of gold = $1764.20 per troy ounce
1 troy ounce = 31.103 g

molarity(M)=molL

molar mass gold (Au) = 196.96 g/mol

processing technique captures only 50 % of the gold present in the samples → multiply mass by 2

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Problem Details

The average concentration of gold in seawater is 100 fM (femtomolar).  Given that the price of gold is $1764.20 per troy ounce (1 troy ounce = 31.103 g), how many liters of seawater would you need to process to collect $7000 worth of gold, assuming your processing technique captures only 50 % of the gold present in the samples? Express your answer using one significant figure.

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