# What mass of potassium hydroxide, KOH, is required to make 250 cm^3 of a 0.5 M solution?

$250$ $c {m}^{3}$ = $0.25$ ${\mathrm{dm}}^{3}$ (= $0.25$ $L$)
For a solution, $C = \frac{n}{V}$ (concentration = number of moles/volume). Rearranging, $n = C V = 0.50 \times 0.25 = 0.125$ $m o l$
We need $0.125$ $m o l$ of KOH and the molar mass of KOH is $56.1$ $g$ $m o {l}^{-} 1$, so $56.1 \times 0.125 = 7.0$ $g$. We need $7.0$ $g$ of KOH.