# How many potassium hydroxide, KOH, formula units are present in 6.89 mol of KOH?

May 29, 2016

$4.15 \cdot {10}^{24}$

#### Explanation:

Your tool of choice here will be Avogadro's number, which can be used as a conversion factor between the number of moles of a given substance and the number of molecules they contain.

In the case of an ionic compound such as potassium hydroxide, $\text{KOH}$, you have formula units instead of molecules.

So, for an ionic compound, Avogadro's number tells you how many formula units you get per mole

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{1 mole" = 6.022 * 10^(23)color(white)(a)"f. units} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ Avogadro's number

In your case, you have to figure out how many formula units of potassium hydroxide you get in $6.89$ moles of this compound.

Use Avogadro's number as a conversion factor to get

6.89 color(red)(cancel(color(black)("moles KOH"))) * overbrace((6.022 * 10^(23)color(white)(a)"f. units")/(1color(red)(cancel(color(black)("mole KOH")))))^(color(blue)("Avogadro's number")) = color(green)(|bar(ul(color(white)(a/a)color(black)(4.15 * 10^(24)color(white)(a)"f units")color(white)(a/a)|)))

The answer is rounded to three sig figs, the number of sig figs you have for the number of moles of potassium hydroxide.