Question

The density of gold is \(19.3 \mathrm{~g} / \mathrm{cm}^{3}\). What volume in milliliters will \(20.0 \mathrm{~g}\) of gold occupy? (Hint: Don't be fooled. Remember that \(1 \mathrm{~cm}^{3}=1 \mathrm{~mL}\).)

Short answer

The volume of 20.0 g of gold is approximately 1.0363 mL.

Step-by-step solution

Calculate the mass of gold in grams

We are given that the mass of gold is 20.0 g.

Use the density formula to find the volume

Density = mass/volume. In our case, we can rearrange the formula to solve for the volume: Volume = mass/density. We are given that the density of gold is 19.3 g/cm³. Calculate the volume using the given mass and density of gold: Volume (in cm³) = \( \frac{20.0 \mathrm{~g}}{19.3 \mathrm{~g}/\mathrm{cm}^{3}} \)

Simplify the expression and calculate the volume

Replace the given values in the expression and simplify: Volume (in cm³) = \( \frac{20.0 \mathrm{~g}}{19.3 \mathrm{~g}/\mathrm{cm}^{3}} \) = \( \frac{20.0 \mathrm{~g}}{1} \times \frac{1 \mathrm{cm}^{3}}{19.3 \mathrm{~g}} \) Volume (in cm³) = \( \frac{20.0}{19.3} \) cm³ = 1.0363 cm³

Convert the volume from cm³ to mL

We are given the conversion factor: 1 cm³ = 1 mL. Volume (in mL) = Volume (in cm³) = 1.0363 mL

State the final answer

The volume of 20.0 g of gold is approximately 1.0363 mL.

Key concepts

Volume Conversion

Converting between different units of volume is a foundational skill in chemistry. In this exercise, we need to convert from cubic centimeters (cm³) to milliliters (mL). Fortunately, this is a straightforward conversion because:

• 1 cm³ is exactly equal to 1 mL.

This direct equivalence allows us to express the calculated volume of gold in cm³ directly as mL without further calculations. It's important to remember this basic conversion factor as it simplifies many volume-related problems in chemistry.

Mass and Volume Relationship

Understanding the relationship between mass and volume is key to solving problems involving density. Mass expresses the amount of substance in grams, while volume is the amount of space that substance occupies, usually measured in cubic centimeters or milliliters.

• The mass of a substance remains constant unless additional material is added or removed.
• Volume, however, can vary depending on the substance's state, temperature, and pressure.

In this exercise, we have a fixed mass of gold and need to find the volume it occupies using its density value. This relationship helps us determine how tightly packed the material is.

Density Formula Application

The density formula is a fundamental concept in chemistry used to relate mass and volume:\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]To solve for volume, rearrange the formula:\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]Applying this formula in the problem, we substitute the density of gold and the given mass to find the volume. Here's what we did:

• Used the density of gold: 19.3 g/cm³.
• Substituted into the formula: \( \text{Volume} = \frac{20.0 \ \text{g}}{19.3 \ \text{g/cm}^3} \).
• Simplified to find the volume: approximately 1.0363 cm³ (or mL).

This systematic approach shows how the density formula allows us to calculate unknown values.

Dimensional Analysis in Chemistry

Dimensional analysis is a powerful tool in chemistry for converting units and checking the consistency of equations. It helps ensure that calculations follow logical unit cancellations and conversions. In the original exercise:

• We started with the equation for volume using known units (g and g/cm³).
• Canceled out grams to ensure the volume was in cm³, which then matched mL.
• Used the conversion 1 cm³ = 1 mL to switch units effectively.

This method highlights the importance of understanding unit relationships and ensures that the result is dimensionally accurate,