Glycerol is a syrupy liquid often used in cosmetics and soaps. A 3.25-L sample of pure glycerol has a mass of 4.10 X 10^3 g.
What is the density of glycerol in g/cm^3?

Question

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kobenhavn kobenhavn · Answer 1 · 2019-08-23T11:18:22+02:00

Answer: The density of glycerol will be [tex]1.26cm^3[/tex]

Explanation:

Density is defined as the mass contained per unit volume.

[tex]Density=\frac{mass}{Volume}[/tex]

Given : Mass of glycerol= [tex]4.10\times 10^3[/tex] g

Volume of bromine = 3.25 L= [tex]3250cm^3[/tex] [tex](1L=1000cm^3)[/tex]

Putting in the values we get:

[tex]Density=\frac{4.10\times 10^3g}{3250cm^3}=1.26g/cm^3[/tex]

Thus density of glycerol will be [tex]1.26cm^3[/tex]

mariejoepagaduan mariejoepagaduan · Answer 2 · 2019-09-06T14:47:23+02:00

The density of glycerol used in cosmetics and soaps is 1.26 g/cm³.

FURTHER EXPLANATION

Density is the measure of the compactness of a material. It is the mass per given volume. This can be expressed mathematically as:

[tex]Density = \frac{mass}{volume}[/tex] (equation 1)

To solve the problem, we first sort the given in the problem:

mass = 4.10 × 10³ g

volume - 3.25 L

The problem is asking for the density in units of g/cm³ which means that we must first convert the volume from liters to cm³.

[tex]volume \ (in \ cm^3) = 3.25 \ L \times \frac{1000 \ mL}{1 \ L} \times \frac{1 \ cm^3}{1 \ mL}\\volume \ (in \ cm^3) = 3250 \ cm^3[/tex]

Now we can plug in the values of mass and volume into equation 1:

[tex]Density = \frac{4.10 \times 10^3 \ g}{3250 \ cm^3}\\\\Density = 1.2615 \frac{g}{cm^3}[/tex]

Since the given in the problem both have 3 significant figures, the final answer must also have three significant digits only. Therefore,

[tex]\boxed {Density = 1.26 \frac{g}{cm^3}}[/tex]

The density of glycerol is greater than 1, so when mixed with water it will sink.

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Keywords: Density, Mass, Volume, Unit Conversion