Answer to Question #239956 in General Chemistry for Sarah

Question

According to estimates, 1.0 L of seawater contains on the average 4.0 x 10-9 g of gold. If gold is worth $390 per troy ounce and a troy ounce weighs 31.103g, how many cubic meters of seawater would have a millions dollars worth of gold in it?

Answer 1

given

1 L Sea water ==> 4 ×10-9 g of gold

390$ per troy ounce

1 troy ounce = 31.103 g

i.e

31.103 g = 1 troy ounce

4 ×10-9 g/L = 4 ×10-9 g/L ×(1 troy ounce/31.103 g)

= 1.286 × 10-10 troy ounce/L

390$ for 1 troy ounce

So

For 1.286 × 10-10 troy ounce/L

1.286 × 10-10 troy ounce × 390$ / 1 troy ounce

= 5.015 × 10-8 $/L

But we want for 106 dollars

==> 106 $/ 5.015 × 10-8 $/L

= 1.99 × 1013 L

1 L = 10-3 m3

1.99 × 1013 L = 1.99 × 1013 L × 10-3 m/L

= 1.99 × 1010 m3 of sea water