The stockroom insurance claims the percent acetic acid in vinegar to it is in $2.0\%$. The thickness of vinegar is $\pu1.106 g mL-1$. Using average molarity given ($\pu0.2844M$) calculation the mass percent acetic acid in vinegar because that comparison come the stockroom claim.

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Not sure how to method the problem but this is what I"ve done:

$$\frac\pu0.2844 mol\pu1L \times (\pu60.05g mol-1)\times \left(\frac\pu1L\pu1000mL\right) = \pu 0.01707822 g mL-1$$

if you have actually $\pu1000mL$, you"ll have actually $\pu0.001106 g\text Acetic Acid \div \pu0.0000170782 g\text Vinegar$ so that will provide $\%$ the acetic acid? I could use part clarification in addressing this


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edited Oct 15 "18 in ~ 18:34
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$1000$ mL that vinegar will certainly weigh $1106$ g and also contain $1000 · 0.017078$ g the acetic mountain (= $17.078$ g). Then $17.078/1106 = 0.0154 = 1.54$% acetic acid.

The stockroom manager should be diluting the vinegar, possibly for a certain experiment. The vinegar i buy is $5$% acetic mountain (rarely $4$%).


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