Using molarity as a conversion factor: An experiment calls for the addition to a reaction vessel of `0.200 g` of sodium hydroxide `(NaOH)` in aqueous solution. How many milliliters of `0.200 M NaOH` should be added?
Strategy: According to equating `2.13`
`M=n_(solute)/V_(L)`
First we need to convert grams `NaOH` to moles `NaOH`, because molarity relates moles of solute to volume of solution. Then, we convert moles `NaOH` to litres of solution, using the molarity as a conversion factor. Here, `0.200 M` means that `1 L` of solution contains `0.200` moles of solute `(NaOH)`, so the conversion factor is:
`(1L soln)/(0.200 mol NaOH)` (Converts mol NaOH to L soln)

Explanation

The formula mass of `NaOH` is 40.0 u formula unit. Thus, the molar mass of `NaOH` is `40.0 g mol^(-1)`. The required calculation is
`Mass rarr "Mole" rarr Litre`
`0.200 g NaOHxx(1 mol NaOH)/(40.0 g NaOH)xx(1 L soln)/(0.200 mol NaOH)`
`=0.025 L soln`
`=25 mL`
We thus need to add `25 mL` of `0.200 M NaOH` solution to the reaction vessel.
Alternatively, according to equation `2.14`
`M=n_(solute)/V_(mL)xx(1000 mL)/L`
`=mass_(solute)/((molar mass)_(solute) V_(mL))xx(1000 mL)/L`
`0.200 M=(0.200 g NaOH)/((40.0 g NaOH)/(mol NaOH)xxV_(mL))xx(1000 mL)/L`
`or V_(mL)=25 mL`