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klemol [59]
3 years ago
14

A 1.0-ml volume of seawater contains about 4.0 3 10212 g of gold. the total volume of ocean water is 1.5 3 1021 l. calculate the

total amount of gold (in grams) that is present in seawater, and the worth of the gold in dollars (see problem 1.82). with so much gold out there, why hasn't someone become rich by mining gold from the ocean?
Chemistry
1 answer:
olga nikolaevna [1]3 years ago
4 0

Given the volume of water = 1.0 mL

Mass of gold = 4.0*10^{-12}g

Total volume of sea water = 1.5 * 10^{21} L

Mass of gold in total sea water

= 1.5 * 10^{21} L*\frac{1000 mL}{1 L} * \frac{4.0*10^{-12}g}{1.0 mL}

=6.0*10^{12}g

1 oz gold costs $948

Worth of gold in dollars = 6.0*10^{12}g*\frac{1 oz}{28.3495 g} *\frac{948 dollars}{1oz}

= $2.0 * 10^{14} /1 oz

Mining of gold from oceans is not practically possible, as gold is spread through out the ocean and processing gold from such vast oceanic waters is not economical and the extraction process is inefficient.

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The isotope that is more abundant, given the data is isotope Li7

<h3>Assumption</h3>
  • Let Li6 be isotope A
  • Let Li7 be isotope B

<h3>How to determine whiche isotope is more abundant</h3>
  • Molar mass of isotope A (Li6) = 6.02 u
  • Molar mass of isotope B (Li7) = 7.02 u
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Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]

6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]

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Cross multiply

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Thus,

Abundance of B = (100 - A)%

Abundance of B = (100 - 8)%

Abundance of B = 92%

SUMMARY

  • Abundance of A (Li6) = 8%
  • Abundance of B (Li7) = 92%

From the above, isotope Li7 is more abundant.

Learn more about isotope:

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