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Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 111
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 111

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# The average concentration of gold in seawater is 100 fM ISBN: 9780321809247 1

## Solution for problem 111 Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

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Problem 111

The average concentration of gold in seawater is 100 fM (femtomolar). Given that the price of gold is \$1764.20 per troy ounce (1 troy ounce = 31.103 g), how many liters of seawater would you need to process to collect \$5000 worth of gold, assuming your processing technique captures only 50% of the gold present in the samples?

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Solution: Here, we are going to calculate the volume of seawater needed to process to collect \$5000 worth of gold. Step1: Given, cost of 1 troy ounce of gold = \$1764.20 i.e., For \$1764.20, quantity of gold collected = 1 troy ounce Therefore, for \$5000, quantity of gold collected = 1 X 5000 [(troy ounce / \$) X \$] 1764.20 = 2.834 troy ounce = 2.834 x 31.103 g [1 troy ounce = 31.103 g] = 88.146 g Step2: Since, the processing technique captures only 50% of the gold present in the sample, therefore, amount of gold needed for processing = (88.146 g / 50 %) x 100 % = 176.292 g Step3: Now, concentration of gold in sea water = 100 fM = 100 x 10 -1mol/L -13 = 10 mol/L Molar mass of gold = 197 g/mol Therefore, number of moles of gold = mass of gold / molar mass = 176.292 / 197 = 0.89 mol Therefore, volume of the sea water in litres = number of moles / molarity -13 = 0.89 mol / (10 mol/L) = 0.89 x 10 L3 = 8.9 x 10 L Thus, required volume of seawater is 8.9 x 10 L. 12

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##### ISBN: 9780321809247

This full solution covers the following key subjects: gold, seawater, Troy, ounce, assuming. This expansive textbook survival guide covers 82 chapters, and 9454 solutions. Since the solution to 111 from 4 chapter was answered, more than 411 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 111 from chapter: 4 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM. The answer to “The average concentration of gold in seawater is 100 fM (femtomolar). Given that the price of gold is \$1764.20 per troy ounce (1 troy ounce = 31.103 g), how many liters of seawater would you need to process to collect \$5000 worth of gold, assuming your processing technique captures only 50% of the gold present in the samples?” is broken down into a number of easy to follow steps, and 58 words. This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3. Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247.

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