Magnesium occurs in seawater to the extent of \(1.4 \mathrm{g}\) magnesium per kilogram of seawater. What volume of seawater, in cubic meters, would have to be processed to produce \(1.00 \times 10^{5}\) tons of magnesium \((1 \text { ton }=2000 \mathrm{lb}) ?\) Assume a density of \(1.025 \mathrm{g} / \mathrm{mL}\) for seawater.

Short Answer

Expert verified
The volume of seawater needed to produce \(1.00 \times 10^{5}\) tons of magnesium is \(1.77 \times 10^{11}\) cubic meters.

Step by step solution

01

Convert tons to grams

In this step, convert the desired amount of magnesium from tons to grams. Note that \(1 \text { ton } = 2000 \text { lb }\) and \(1 \text { lb } = 453.592 \text { g }\). So you multiply \(1.00 \times 10^{5}\) tons by 2000 and then by 453.592 to get the result in grams.
02

Calculate the amount of seawater needed

After converting the desired amount of magnesium to grams, you can now calculate the required amount of seawater. Given that 1.4g of magnesium is present per kg of seawater, you divide the total amount of magnesium (in grams) by 1.4 to get the required amount of seawater in kg.
03

Convert kilograms to cubic meters

As the density of seawater is given as \(1.025 \text { g/mL }\) and we know that \(1 \text { kg } = 1000 \text { g }\) and \(1 \text { mL } = 0.001 \text { m$^3$ }\), we can easily convert the required amount of seawater from kg to cubic meters by multiplying by \( \frac{1}{1.025} \) and then by 0.001.

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