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A vinegar sample is found to accept a density of $1.006 \mathrm{\frac{g}{mL}}$ and to comprise $8.vii~\%$ acetic acid by mass. How many grams of acetic acid are present in $\pu{2.80L}$ of this vinegar?

I multiplied: $$\left(\frac{ane.006~\mathrm{one thousand}}{\mathrm{ml}}\right)\cdot \left(\frac{1000~\mathrm{ml}}{1~\mathrm{L}}\right)\cdot 2.80~\mathrm{L} = 2816.8~\mathrm{g}$$

I so converted $8.7~\%$ into $0.087$ and multiplied it into $2816.8$, to get the answer $245~\mathrm{g}$.

Am I right?

  • physical-chemical science
  • concentration

Jan's user avatar

Jan

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asked Sep 12, 2014 at 21:09

Cetshwayo's user avatar

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5

  • $\begingroup$ Delight add what you have attempted towards solving the problem into the trunk of your question. For more information, see the site'south homework policy for how to ask homework questions. Thank you! $\endgroup$

    Sep 12, 2014 at 21:13

  • $\begingroup$ (you've done this very well in your other questions so far, then if you lot could do the same with this one, that would exist great) $\endgroup$

    Sep 12, 2014 at 21:xviii

  • $\begingroup$ Can yous tell me if I am on the right runway? $\endgroup$

    Sep 12, 2014 at 22:19

  • $\begingroup$ Yes, I get the same answer every bit yous. $\endgroup$

    Sep 12, 2014 at 22:45

  • $\begingroup$ @JohnSnow I appreciate you working hard in editing the questions with MathJax. I would appreciate it even more, if you would take a wait at this meta post, specifically the department about units and variables. $\endgroup$

    Sep 13, 2014 at vi:41

1 Answer 1

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Your math and stoichiometry are correct. $2.80L*(\frac{chiliad ml}{1 50})*(\frac{1.006 grand}{ ml})*eight.7$%$=245g \checkmark$

answered Sep 13, 2014 at 1:05

John Snow's user avatar

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