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

1 Answer
May 29, 2016

4.15 * 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, "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

color(blue)(|bar(ul(color(white)(a/a)"1 mole" = 6.022 * 10^(23)color(white)(a)"f. units"color(white)(a/a)|))) -> 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.