How many Au atoms can be extracted from 0.250 L of seawater? 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?
You also need to convert the grams of gold into moles in order to use Avogadro's number.

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