# Which solution contains 0.2 mol of potassium hydroxide, KOH? (multiple choice)

## A. 1 $c {m}^{3}$ of 0.2 mol ${\mathrm{dm}}^{3}$ KOH B. 10 $c {m}^{3}$ of 0.2 mol ${\mathrm{dm}}^{3}$ KOH C. 100 $c {m}^{3}$ of 2.0 mol ${\mathrm{dm}}^{3}$ KOH D. 1 ${\mathrm{dm}}^{3}$ of 2.0 mol ${\mathrm{dm}}^{3}$ KOH The answer is C. 100 $c {m}^{3}$ of 2.0 mol ${\mathrm{dm}}^{3}$ KOH, why?

Feb 20, 2016

$\text{Concentration}$ $=$ $\text{Number of moles (n)"/"Volume of solution (L)}$

#### Explanation:

Thus, since $\text{Concentration} = \frac{n}{V}$, all we have to do is multiply the concentration by the volume to get the number of moles: we can even do so dimensionally, i.e. $\text{Concentration "xx" Volume}$ $=$ $m o l \cdot \cancel{{L}^{-} 1} \times \cancel{L}$ $=$ $m o l$ as required.

$0.100 \cdot L \times 2.0 \cdot m o l \cdot {L}^{-} 1 = 0.200 \cdot m o l$.

Note that $1000$ $c {m}^{3}$ $=$ $1$ $L$ or $1$ ${\mathrm{dm}}^{3}$. If I expanded out ${\mathrm{dm}}^{3}$, I would get ${\left({10}^{- 1} m\right)}^{3}$, $=$ ${10}^{- 3} {m}^{3}$. Or $1000$ $L$ $=$ $1$ ${m}^{3}$.

You would not be asked to perform this conversion at A level; in 1st year chemistry I would expect it. You WOULD be expected to know that $1000$ $c {m}^{-} 3$ $=$ $1$ ${\mathrm{dm}}^{3}$ $=$ $1 \cdot \text{L}$. Use what unit you like, but be consistent. This is of course a lot of work to do for a single multiple choice question.