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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

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  • $\begingroup$ HINT: You want: mass(vinegar)/mass(solution) $\endgroup$ – MaxW Oct 15 '18 at 20:42
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1000 mL of vinegar will weigh 1106 g and contain 1000 x 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%).

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