A1.0 mL sample of seawater contains 0.00000000004g of gold the total volume of seawater is 150000000000000000000L calculate the total mass of gold present in seawater?

Answer 1

based on your assumed data:

4.0*10-11 gms/ml, 1000ml per L => 4.0*10-8 gms/L => 4.0*10-5 grams/1000L or 1 tonne

15.0*1019 L

4*15==60=6.0*101

1019 L *10-8 gms/L *101 => 19-8+1=12 adding powers

Answer:

6.0 *1012 gms.

or 6 *109Kgms = 6,000,000,000 kgms

Your data may be showing too low a concentration of gold.

"It's been estimated that there is a milligram of gold in a ton of seawater."

One tonne (metric) tonne is 1000 litres.

4.0*10-5 grams/1000L = 4.0*10-2 milligrams/1000L = 0.04 milligrams/tonne

The quote above, if correct is 1.0 milligrams/tonne or 25 times what you are showing.

----some clippings on amount of gold in seawater - I don't know if these are accurate--

There is a huge quantity of gold is seawater. But it's distributed widely - across all the oceans of all the world - and is incorporated on the atomic level. That makes it very, very hard to separate out. No one has demonstrated that it can be done economically.

The concentration of gold in seawater varies from place to place, and ranges between 5 to 50 ppt (about .005 to .05 tonnes (5 to 50 kg) per km3)

{It's been estimated that there is a milligram of gold in a ton of seawater.}

Given that the volume of all the seas is about 1.37 billion km3 The total amount of gold in all the seas is about 7 to 75 billion kg.

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Gold occurs in sea water to the extent of 0.1 to 2 mg/ton, depending on the location where the sample is taken.

1 cubic metre is equivalent to:

* 1,000 litres

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A cubic metre of pure water at the temperature of maximum density (3.98 °C) and standard atmospheric pressure (101.325 kPa) has a mass of 1000 kg, or one tonne.