   # A sample of an ethanol–water solution has a volume of 54.2 cm 3 and a mass of 49.6 g. What is the percentage of ethanol (by mass) in the solution? (Assume that there is no change in volume when the pure compounds are mixed.) The density of ethanol is 0.789 g/cm 3 and that of water is 0.998 g/cm 3 . Alcoholic beverages are rated in proof , which is a measure of the relative amount of ethanol in the beverage. Pure ethanol is exactly 200 proof; a solution that is 50% ethanol by volume is exactly 100 proof. What is the proof of the given ethanol–water solution? ### General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
Publisher: Cengage Learning
ISBN: 9781305580343

#### Solutions

Chapter
Section ### General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
Publisher: Cengage Learning
ISBN: 9781305580343
Chapter 1, Problem 1.169QP
Textbook Problem
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## A sample of an ethanol–water solution has a volume of 54.2 cm3 and a mass of 49.6 g. What is the percentage of ethanol (by mass) in the solution? (Assume that there is no change in volume when the pure compounds are mixed.) The density of ethanol is 0.789 g/cm3 and that of water is 0.998 g/cm3. Alcoholic beverages are rated in proof, which is a measure of the relative amount of ethanol in the beverage. Pure ethanol is exactly 200 proof; a solution that is 50% ethanol by volume is exactly 100 proof. What is the proof of the given ethanol–water solution?

Interpretation Introduction

Interpretation:

The percentage of ethanol by mass in the given solution has to be calculated and the proof of given ethanol-water solution has to be given.

Concept Introduction:

Mass per unit volume of any object is known as its density.  Density can be calculated if mass and volume of the object is known.  The SI unit of density is kg/m3 .

d = mV

Where,

d is the density

m is the mass

V is the volume.

### Explanation of Solution

The total mass of the solution is given as 49.6g and volume is 54.2cm3 .

Let us consider mass of ethanol be “x” and mass of water be “y”.

Therefore,

x+y =   49.6g .

From the above equation, the mass of water is 49.6g-x .

Total volume of solution is equal to the sum of volume of ethanol and volume of water.

Totalvolume = mass of ethanoldensity of ethanol + mass of waterdensity of water

Substituting the values in the above equation, we get

54.2 cm3 = x0.789 g/cm3 + 49.6 g - x0.998 g/cm3

Rearranging the above equation we get,

(0.789)(0.998)(54.2) g = 0.998x + 0.789 (49.6 g - x)42.678 g = 0.998 x  +  39.134 g - 0.789 x = 0.209 x  + 39.134 g0.209 x = 42.678 g  -  39.134 g = 3.544 gx = 3.544 g0.209 = 16.95g

Therefore, the mass of ethanol is 16.95g .

The percentage by mass of ethanol in the solution can be calculated as shown below,

Percent(mass) = mass of ethanolmass of solution×100% = 16

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