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Gold is present in seawater to the extent of 0.15 mg/ton. Assume the density of seawater is 1.03 g/ml and determine how many gold atoms could be extracted from 0.250 L of seawater.

($\pu{1 ton}=\pu{2000lbs}, \pu{1kg}=\pu{2.205lbs}, \pu{1 mole}=\pu{6.022E23 atoms}, \text{1 mole of gold}=\pu{197g}$)

I just multiplied everything together using dimensional analysis and got $5.2\times10^{14}$ atoms of Au per liter of seawater, but they want how many atoms of Au in $\pu{0.250L}$ of seawater so I just multiplied $5.2\times10^{14}$ by $0.250$ to get $1.30\times10^{14}$ atoms of Au per $\pu{0.250L}$ of seawater. Can you simply multiply everything together like this? When can you do this and when can't you?

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    $\begingroup$ Note that since the rest of the units in the problem are SI, there's a good chance that "ton" is supposed to be a metric ton, which is 1000 kg instead of 2000 lb. $\endgroup$ – Nate Eldredge Sep 15 '14 at 1:02
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You also need to convert the grams of gold into moles in order to use Avogadro's number.

Atomic weight of gold = $\pu{196.967 g/mol}$ $$ \begin{align} \frac{\pu{1.5E-04 g/ton}}{\pu{196.967 g/mol}}&= \pu{7.615E-07 mol/ton}\\ \pu{7.615E-07 mol/ton}\times \pu{6.022E+23 atoms/mol} &= \pu{4.586E+17 atoms/ton}\\ \frac{\pu{4.586E{+17} atoms/ton}}{\pu{2000 lb/ton}}&=\pu{2.293E{+14} atoms/lb}\\ \frac{\pu{2.293E{+14} atoms/lb}}{\pu{0.454 kg/lb}}=\pu{5.056E{+14} atoms/kg}&=\pu{5.056E{+11} atoms/g}\\ \pu{5.056E{+11} atoms/g} \times \pu{1.03 g/ml}&=\pu{5.208E{+11} atoms/ml}\\ \pu{5.208E{+11} atoms/ml}\times \pu{250 ml}&= \pu{1.302 E{+14} atoms}/\pu{250 ml}\\ \end{align} $$ There would be $1.302 \times10^{14}$ atoms of gold in $\pu{0.250 L}$ of seawater.

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