Please help with a Concentration equation?

Question

The question is:
Gold occurs in seawater at an average concentration of 1.1X10^-2 parts per billion. How many liters of seawater must be processed to recover 31.1 g of gold assuming 79.5% efficiency (density of seawater = 1.025 g/mL)

any help would be great! If you could explain it and help me to get it on my own even better.

Answer 1

OK, lets try...to get 31.1 g of gold if the process is only 79.5% efficient, we would need to try and get:

31.1 g/X g = 79.9%/100%
X g = 31.1 g /0.795
X g = 39.1 g

To get the amount of seawater we can use a similar way:
(P.S. this is with a US billion, that is 1000000000)

39.1 g/X g = 0.0011/1000000000
X g = (39.1 g)(1000000000)/0.0011
X g = 3.55 X 10^13 g of water

Now lets work on volume:

(3.55 X 10^13 g)(1 mL/1.025 g)(1 L/1000 mL)
3.47 X10^10 L of seawater

Write these things out so you can see the algebra and how the units cancel. Good Luck