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Chapter 1: Problem 56

Magnesium (Mg) is a valuable metal used in alloys, in batteries, and in chemical synthesis. It is obtained mostly from seawater, which contains about \(1.3 \mathrm{~g}\) of Mg for every kilogram of seawater. Calculate the volume of seawater (in liters) needed to extract \(8.0 \times 10^{4}\) tons of \(\mathrm{Mg}\), which is roughly the annual production in the United States. (Density of seawater = \(1.03 \mathrm{~g} / \mathrm{mL} .)\)

Short Answer

Expert verified
You would need approximately \(5.97 \times 10^{10}\: \mathrm{L}\) of seawater to produce \(8.0 \times 10^{4}\) tons of Magnesium per year.

Step by step solution

01

Convert tons of Mg to grams

Given that \(1\) ton equals \(10^{6} \: \mathrm{g}\), we multiply \(8.0 \times 10^{4}\) tons by \(10^{6} \: \mathrm{g}/\mathrm{ton}\) to get \(\: 8.0 \times 10^{4} \times 10^{6} = 8.0 \times 10^{10}\: \mathrm{g}\) of Magnesium.
02

Determine the volume of seawater

Given that \(1.3 \: \mathrm{g}\) of Mg can be obtained from \(1 \: \mathrm{kg}\) (or \(1000 \: \mathrm{g}\)) of seawater, we set up a proportion to find the mass of seawater required to extract \(8.0 \times 10^{10}\: \mathrm{g} \) of Mg. This is done by solving the proportion \(\frac{1.3}{1000} = \frac{8.0 \times 10^{10}}{x}\) for \(x\), which results in \(x = \frac{8.0 \times 10^{10} \times 1000}{1.3} = 6.15 \times 10^{13}\: \mathrm{g}\) of seawater.
03

Convert grams of seawater to liters

Knowing that the density of seawater is \(1.03\: \mathrm{g/mL}\), we can use this to convert grams of seawater to mL (and then to liters). \(\frac{6.15 \times 10^{13} \: \mathrm{g}}{1.03 \: \mathrm{g/mL}} = 5.97 \times 10^{13}\: \mathrm{mL}\) of seawater. Since \(1\: \mathrm{L} = 1000\:\mathrm{mL}\), the volume in liters is \(\frac{5.97 \times 10^{13}}{1000} = 5.97 \times 10^{10}\: \mathrm{L}\)
04

Answer

The volume of seawater (in liters) needed to extract \(8.0 \times 10^{4}\) tons of magnesium is roughly \(5.97 \times 10^{10}\: \mathrm{L}\)

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