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Chapter 13: Problem 15

A certain vinegar is \(6.02 \%\) acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) by mass. How many grams of \(\mathrm{CH}_{3} \mathrm{COOH}\) are contained in a \(355 \mathrm{mL}\) bottle of vinegar? Assume a density of \(1.01 \mathrm{g} / \mathrm{mL}\).

Short Answer

Expert verified
There are approximately 21.6 grams of acetic acid in the 355 mL bottle of vinegar.

Step by step solution

01

Convert volume to mass

Using the density, one can convert the volume of the vinegar to its mass. The density, tells us that in each mL there are 1.01 g vinegar. So, if we have 355 mL of vinegar, we can multiply this volume by the density to find the mass: \(355 \, mL \, \times 1.01 \, g/mL\) which equals \(358.55 \, g\)
02

Calculate mass of acetic acid

We know that the vinegar is 6.02% acetic acid by mass. This means that for every 100 g of vinegar, there are 6.02 g of acetic acid. Therefore, for \(358.55 \, g\) of vinegar, we can multiply this by \(6.02/100\) to find the mass of the acetic acid in the vinegar: \(358.55 \, g \times (6.02 / 100) = 21.58 \, g\)
03

Round off the calculation

The above result is not rounded since calculations were being performed. Now, it is time to report the final result. Following significant figure rules which state that your answer should not have more significant figures than the least accurately known number, in this case, the percentage given to 3 significant figures, we get the answer: \(21.6 \, g\)

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Most popular questions from this chapter

An ideal liquid solution has two volatile components. In the vapor in equilibrium with the solution, the mole fractions of the components are (a) both \(0.50 ;\) (b) equal, but not necessarily \(0.50 ;\) (c) not very likely to be equal; (d) 1.00 for the solvent and 0.00 for the solute.

The solubility of a nonreactive gas in water increases with (a) an increase in gas pressure; (b) an increase in temperature; (c) increases in both temperature and pressure; (d) an increase in the volume of gas in equilibrium with the available water.

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