The bromine content of the ocean is about 65 grams of bromine per million grams of sea water. How many cubic

meters of ocean must be processed to recover 1.0 pounds of bromine if the density of sea water is 1.0 x 103 kg/m3?

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1. 65 ppm is 65 grams/10^6 g sea water. We want 1 lb of Br2 which is about 454 grams. So how much sea water do we need to do that?
That's 1E6 g sea water x (454/65) = about 7E6 grams. I would change that to 7E3 kg and use the density.
volume = mass/density = 7E3/1E3 = aboaut 7 cubic meters. You need to go through and clean up the numbers because I've estimated here and there although I think the 7 is close. Check my thinking.

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2. what the heck are you even doing

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3. I got 67.75m cubed.

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4. I got 6.99 m^3. It's

1 lb Br/1
1,000,000 lb Water/65 lb Br
1 kg Water/2.2 lb Water
m^3 Water/1000 kg Water
(Multiply (1)*(1,000,000)*(1)*(m^3))
&
(Multiply (1)*(65)*(2.2)*(1,000))
Take the first answer (1,000,000 m^3)
and divide by the second answer (143,000)
to get 6.99 m^3. Enjoy!

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