Question

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

Answer 1

One mole of any compound or element always contains the same number of units or particles, called Avogadro number. This value is $6.022\times {{10}^{23}}$ particles or formula units.

Complete step-by-step answer: Potassium hydroxide is an ionic compound. The exact numbers of molecules in the empirical formula of ionic compounds are termed as formula units. The molar mass of all the species in a compound multiplied by their amount or quantity in that compound gives us the formula unit mass.
The empirical formula of a compound tells us about the formula units. For instance, in potassium hydroxide, one potassium ion is needed for one hydroxide ion always.
As we know 1 mole of any compound contains $6.022\times {{10}^{23}}$formula units, so 1 mole of KOH will also contain $6.022\times {{10}^{23}}$formula units.
Now to calculate formula units in 6.89 moles of KOH, we will use stoichiometric calculations and conversion factor in terms of mole and multiply the given moles with the formula units in 1 mole of KOH, as:
Formula units in 6.89 moles of KOH = 6.89 moles $\times \dfrac{6.022\times {{10}^{23}}\,formula\,units}{1\,mol\,KOH}$
Formula units in 6.89 moles of KOH = $4.15\times {{10}^{24}}$ formula units
Hence, in 6.89 moles of potassium hydroxide, KOH, $4.15\times {{10}^{24}}$ formula units are present.

Note: Stoichiometric calculations need to be done carefully by employing correct unitary method and conversion factors. Here, only the number of moles of potassium hydroxide is given, so molar mass of potassium hydroxide need not be involved, so simple unitary calculation is done by converting given moles into formula units.