CHEMISTRY HIGH SCHOOL

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?

Answers

Answer 1

Answer:

Answer: The density of glycerol will be $1.26cm^3$

Explanation:

Density is defined as the mass contained per unit volume.

$Density=(mass)/(Volume)$

Given : Mass of glycerol= $4.10* 10^3$ g

Volume of bromine = 3.25 L= $3250cm^3$ $(1L=1000cm^3)$

Putting in the values we get:

$Density=(4.10* 10^3g)/(3250cm^3)=1.26g/cm^3$

Thus density of glycerol will be $1.26cm^3$

Answer 2

Answer:

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:

$Density = (mass)/(volume)$ (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.

$volume \ (in \ cm^3) = 3.25 \ L * (1000 \ mL)/(1 \ L) * (1 \ cm^3)/(1 \ mL)\nvolume \ (in \ cm^3) = 3250 \ cm^3$

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

$Density = (4.10 * 10^3 \ g)/(3250 \ cm^3)\n\nDensity = 1.2615 (g)/(cm^3)$

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

$\boxed {Density = 1.26 (g)/(cm^3)}$

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

LEARN MORE

Learn more about Displacement Method brainly.com/question/11474085
Learn more about Liquid Density brainly.com/question/11301218
Learn more about Dimensional Analysis brainly.com/question/1557970

Keywords: Density, Mass, Volume, Unit Conversion

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Answers

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Explanation:

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