# Question c66da

Approx. $3 \times {10}^{-} 3$ $m o l \cdot {L}^{-} 1$
$\text{Concentration, molarity}$ $=$ $\text{Moles of solute"/"Volume of solution}$ $=$ $\frac{1.56 \cdot m o l}{500 \cdot L}$ $=$ ??# $m o l \cdot {L}^{-} 1$.
By doing the calculation this way and including the units, I am introducing a reality check on my calculation. Do the resultant units make sense? Well, they are not $L \cdot m o {l}^{-} 1$ but $m o l \cdot {L}^{-} 1$, as is required for a concentration.