# How do I calculate the molarity of 7.24 * 10^2 mL of solution containing 22.4 g of potassium iodide?

Jun 30, 2017

$\text{Molarity"="Moles of solute"/"Volume of solution}$.........

$= 0.186 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

And thus.......

$\text{Molarity} = \frac{\frac{22.4 \cdot g}{166.0 \cdot g \cdot m o {l}^{-} 1}}{7.24 \times {10}^{2} \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1}$

$= 0.186 \cdot m o l \cdot {L}^{-} 1$

So what's unusual about this problem? It expresses volume in units of $m L$ NOT $L$. But the prefix $m$ stands for $\text{milli}$ i.e. ${10}^{-} 3$, just as $c$ stands for $\text{centi}$ i.e. ${10}^{-} 2$, and $k$ stands for $\text{kilo}$ i.e. ${10}^{3}$, and $\mu$ stands for $\text{micro}$ i.e. ${10}^{-} 6$. These must be simply be learned, but use them first in practice problems.

And so when we are quoted a volume of $7.24 \times {10}^{2} \cdot m L$, we make the conversion......

$7.24 \times {10}^{2} \cdot m L \equiv 7.24 \times {10}^{2} \cdot \cancel{m L} \times {10}^{-} 3 \cdot L \cdot \cancel{m {L}^{-} 1} = 0.724 \cdot L$

Such operations are known as $\text{dimensional analysis}$, and we did a little example in the problem; we wanted an answer with units of concentration and we got one. It is certainly not straightforward and intuitive, but it does work (eventually!). At your level you simply have to remember that $1 \cdot L = 1 \cdot {\mathrm{dm}}^{3} = 1000 \cdot m L = 1000 \cdot c {m}^{3}$. As you do more problems the conversion becomes a bit more familiar.