Acetic Acid % in vinegar? Given average molarity of acetic acid Question:


  The stockroom claims the percent acetic acid in vinegar to be $2.0\%$. The density of vinegar is $\pu{1.106 g {mL}-1}$. Using average molarity given ($\pu{0.2844M}$) calculate the mass percent acetic acid in vinegar for comparison to the stockroom claim.


Not sure how to approach the problem but this is what I've done:

$$\frac{\pu{0.2844 mol}}{\pu{1L}} \times (\pu{60.05g {mol}-1})\times \left(\frac{\pu{1L}}{\pu{1000mL}}\right) = \pu{ 0.01707822 g {mL}-1}$$

if you have $\pu{1000mL}$, you'll have $\pu{0.001106 g}\text{ Acetic Acid} \div \pu{0.0000170782 g}\text{ Vinegar}$ so that will give $\%$ of acetic acid? I could use some clarification in solving this
$1000$ mL of vinegar will weigh $1106$ g and contain $1000 · 0.017078$ g of acetic acid (= $17.078$ g). Then $17.078/1106 = 0.0154 = 1.54$% acetic acid.
The stockroom manager must be diluting the vinegar, perhaps for a specific experiment. The vinegar I buy is $5$% acetic acid (rarely $4$%).