Problem 70
Question
A student wonders whether a piece of jewelry is made of pure silver. She determines that its mass is \(3.17 \mathrm{g} .\) Then she drops it into a 10 -mL graduated cylinder partially filled with water and determines that its volume is \(0.3 \mathrm{mL}\). Could the jewelry be made of pure silver?
Step-by-Step Solution
Verified Answer
Explain your answer.
1Step 1: Find the density of the material
To find the density of the material we will use the formula:
Density = \(\frac{Mass}{Volume}\)
Given the mass and volume of the jewelry, we can find its density.
2Step 2: Calculate the density of the jewelry
We have the mass of the jewelry as \(3.17 \mathrm{g}\) and its volume as \(0.3 \mathrm{mL}\). Substitute these values into the density formula.
Density = \(\frac{3.17}{0.3} = 10.57 \mathrm{\frac{g}{mL}}\)
So, the density of this jewelry is \(10.57 \mathrm{\frac{g}{mL}}\).
3Step 3: Compare the density with the density of pure silver
The density of pure silver is \(10.49 \mathrm{\frac{g}{mL}}\). Compare this value with the calculated density of the jewelry, which is \(10.57 \mathrm{\frac{g}{mL}}\).
We can see that the density of the jewelry is slightly higher than the density of pure silver. This means that the jewelry could contain a metal with a higher density than silver or that it is an alloy.
4Step 4: Conclusion:
Based on the comparison between the density of the jewelry and the density of pure silver, we cannot conclusively say that the jewelry is made of pure silver. It could be an alloy or a different material altogether.
Key Concepts
MassVolumeDensity of Silver
Mass
When discussing the concept of mass, we are referring to the amount of matter within an object. In the context of our example, the mass of the jewelry in question is given as \(3.17\, \text{g}\). Mass is usually measured in grams (g) or kilograms (kg) depending on the context.
Understanding mass is crucial because it is a fundamental measurement in physics and chemistry.
Understanding mass is crucial because it is a fundamental measurement in physics and chemistry.
- Mass is consistent regardless of location. This means that the mass of the piece of jewelry would remain \(3.17\, \text{g}\) whether on Earth or in space.
- Mass is often confused with weight, but they are not the same. Weight is the force exerted by gravity on an object.
Volume
Volume is the amount of space that an object occupies. For our jewelry piece, its volume is determined to be \(0.3\, \text{mL}\). Volume can be measured in various units like cubic centimeters (cm³), milliliters (mL), or liters (L), depending on the object.
Seeing volume in a new light helps when deciphering how matter surrounds us.
Seeing volume in a new light helps when deciphering how matter surrounds us.
- Accurate measurement of volume is crucial when calculating density.
- In our example, a graduated cylinder was used to find the displacement caused by the jewelry - a common method for irregular objects.
Density of Silver
The concept of density combines both mass and volume to describe how compact an object is. It is given by the formula \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). For pure silver, its known density is \(10.49\, \text{g/mL}\).
In the exercise, by calculating the density of the jewelry as \(10.57\, \text{g/mL}\), we can draw comparisons:
In the exercise, by calculating the density of the jewelry as \(10.57\, \text{g/mL}\), we can draw comparisons:
- Density measures how much mass is packed into a given volume.
- This small difference suggests the jewelry might not be pure silver, as other metals could be present. This would change the calculated density.
Other exercises in this chapter
Problem 67
What volume of gold would be equal in mass to a piece of copper with a volume of \(125 \mathrm{cm}^{3} ?\)
View solution Problem 69
What is the volume of \(1.00 \mathrm{kg}\) of mercury?
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The average density of Earth is \(5.5 \mathrm{g} / \mathrm{cm}^{3} .\) The mass of Venus is \(81.5 \%\) of Earth's mass, and the volume of Venus is \(88 \%\) of
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The sun is a sphere with an estimated mass of \(2 \times 10^{30} \mathrm{kg} .\) If the radius of the sun is \(7.0 \times 10^{5} \mathrm{km}\) what is the avera
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