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

1 Answer
May 29, 2016


#4.15 * 10^(24)#


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.