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Chemistry / Principles of General Chemistry 2 / Chapter 1 / Problem 71P

Principles of General Chemistry | 2nd Edition | ISBN: 9780073511085 | Authors: Martin S. Silberberg

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Earth’s oceans have an average depth of 3800 m, a total

Chapter 2, Problem 71P

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QUESTION:

Earth’s oceans have an average depth of 3800 m, a total area of \(3.63\times10^8\ \text{km}^2\), and an average concentration of dissolved gold of \(5.8\times10^{-9}\ \text{g/L}\). (a) How many grams of gold are in the oceans? (b) How many m3 of gold are in the oceans? (c) Assuming the price of gold is $370.00/troy oz. What is the value of gold in the oceans (1 troy oz = 31.1 g; d of gold = \(19.3\ \text{g/cm}^3\))?

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QUESTION:

Earth’s oceans have an average depth of 3800 m, a total area of \(3.63\times10^8\ \text{km}^2\), and an average concentration of dissolved gold of \(5.8\times10^{-9}\ \text{g/L}\). (a) How many grams of gold are in the oceans? (b) How many m3 of gold are in the oceans? (c) Assuming the price of gold is $370.00/troy oz. What is the value of gold in the oceans (1 troy oz = 31.1 g; d of gold = \(19.3\ \text{g/cm}^3\))?

ANSWER:

Step 1 of 3

(a)

From the given,

Surface area =

Let’s convert the surface area “km” into “m”.

Depth = 3800m

Concentration = .

Let’s calculate the number of grams of gold in the ocean.

Therefore, the number of grams of gold in the ocean is

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