Suppose that I am trying to determine if a piece of rock is either "fool's gold" or pure gold. The density of pure gold is 19.3 g/mL, and I decide to use the water displacement method. The rock has a mass of 193 g, and I start with a volume of water at 20 mL. What should the final water level be after placing the rock into the container for it to be considered pure gold? O 25 mL O 30 mL 35 mL 40 ml m DV X
Suppose that I am trying to determine if a piece of rock is either "fool's gold" or pure gold. The density of pure gold is 19.3 g/mL, and I decide to use the water displacement method. The rock has a mass of 193 g, and I start with a volume of water at 20 mL. What should the final water level be after placing the rock into the container for it to be considered pure gold? O 25 mL O 30 mL 35 mL 40 ml m DV X
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![**Determining the Density of a Rock to Identify as Gold or Fool's Gold Using Water Displacement**
In this exercise, we aim to determine if a rock is either "fool's gold" or pure gold using the water displacement method. The density of pure gold is 19.3 g/mL. Given:
- Mass of the rock: 193 g
- Initial volume of water: 20 mL
**Goal:** To find the final water level needed to consider the rock as pure gold.
**Formula for Density:**
\[ \text{Density} (D) = \frac{\text{Mass} (m)}{\text{Volume} (v)} \]
**Steps to Solve:**
1. Compute the volume \( v \) using the formula rearranged from the density equation:
\[ v = \frac{m}{D} = \frac{193 \, \text{g}}{19.3 \, \text{g/mL}} \approx 10 \, \text{mL} \]
2. Add the volume of the rock (10 mL) to the initial volume of water (20 mL):
\[ 20 \, \text{mL} + 10 \, \text{mL} = 30 \, \text{mL} \]
**Conclusion:** The final water level should be 30 mL for the rock to be considered pure gold.
**Explanation of Diagram:**
The diagram shows a triangle with:
- \( m \) at the top, representing mass
- \( D \) and \( v \) at the bottom corners, representing density and volume respectively
- The arrangement reminds users of the formula \( D = \frac{m}{v} \)
**Multiple Choice Options:**
- 25 mL
- **30 mL** (Correct Answer)
- 35 mL
- 40 mL
This method helps verify the purity of materials based on density analysis.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F58c20e32-c100-4a01-973d-82f8e2d65138%2Fb56ae40e-49e0-4b7d-9238-16a734c1e36a%2Fux4u0a4_processed.png&w=3840&q=75)
Transcribed Image Text:**Determining the Density of a Rock to Identify as Gold or Fool's Gold Using Water Displacement**
In this exercise, we aim to determine if a rock is either "fool's gold" or pure gold using the water displacement method. The density of pure gold is 19.3 g/mL. Given:
- Mass of the rock: 193 g
- Initial volume of water: 20 mL
**Goal:** To find the final water level needed to consider the rock as pure gold.
**Formula for Density:**
\[ \text{Density} (D) = \frac{\text{Mass} (m)}{\text{Volume} (v)} \]
**Steps to Solve:**
1. Compute the volume \( v \) using the formula rearranged from the density equation:
\[ v = \frac{m}{D} = \frac{193 \, \text{g}}{19.3 \, \text{g/mL}} \approx 10 \, \text{mL} \]
2. Add the volume of the rock (10 mL) to the initial volume of water (20 mL):
\[ 20 \, \text{mL} + 10 \, \text{mL} = 30 \, \text{mL} \]
**Conclusion:** The final water level should be 30 mL for the rock to be considered pure gold.
**Explanation of Diagram:**
The diagram shows a triangle with:
- \( m \) at the top, representing mass
- \( D \) and \( v \) at the bottom corners, representing density and volume respectively
- The arrangement reminds users of the formula \( D = \frac{m}{v} \)
**Multiple Choice Options:**
- 25 mL
- **30 mL** (Correct Answer)
- 35 mL
- 40 mL
This method helps verify the purity of materials based on density analysis.
Expert Solution
Step 1
Given data :
Density of pure gold = 19.3 g.mL-1
Mass of rock = 193 g
Initial volume of water = 20 mL
Final volume of water = ?
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