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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) |
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Answer» 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` |
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