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Chapter 1: Problem 54

A vinegar sample is found to have a density of \(1.006 \mathrm{g} / \mathrm{mL}\) and to contain \(5.4 \%\) acetic acid by mass. How many grams of acetic acid are present in \(1.00 \mathrm{L}\) of this vinegar?

Short Answer

Expert verified
The mass of acetic acid present in 1.00 L of this vinegar is 54.324 grams.

Step by step solution

01

Calculate Total Mass

The density of a substance is defined as its mass per unit volume. Therefore, for the solution, we have: Mass = Density * Volume. Given, Density of Vinegar = 1.006 g/mL and Volume = 1.00 L = 1000 mL (since 1L = 1000mL), Thus, Total mass of the solution = 1.006 g/mL * 1000 mL = 1006 g.
02

Find Mass of Acetic Acid

Next, we find the mass of the acetic acid. Acetic acid makes up 5.4% of the vinegar by mass. Therefore, the mass of the acetic acid in 1006 g of vinegar can be calculated as follows: Mass of Acetic Acid = 5.4% of 1006 g = 0.054 * 1006 g = 54.324 g.

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