Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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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.
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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
Check Mark
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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