Problem 81 - Chemistry The Central Science

Question

Water has a density of 0.997 \(\mathrm{g} / \mathrm{cm}^{3}\) at \(25^{\circ} \mathrm{C} ;\) ice has a density of 0.917 \(\mathrm{g} / \mathrm{cm}^{3}\) at \(-10^{\circ} \mathrm{C}\) (a) If a soft-drink bottle whose volume is 1.50 \(\mathrm{L}\) is completely filled with water and then frozen to \(-10^{\circ} \mathrm{C},\) what volume does the ice occupy? (b) Can the ice be contained within the bottle?

Step-by-Step Solution

Answer

The mass of the water in the bottle is approximately 1495.5 g. When frozen, the ice occupies a volume of approximately 1630.4 cm³. Since the volume of the ice is greater than the volume of the bottle (1500 cm³), the ice cannot be contained within the bottle.

Step 1: Calculate the mass of the water

First, we need to find the mass of the water in the bottle before freezing. We can do this by using the formula for finding mass, which is density times volume. We are given the volume of the bottle, 1.50 L, and the density of the water, 0.997 g/cm³. First, we need to convert the volume from liters to cubic centimeters, as follows: \[1.50\, \text{L} \times \frac{1000\, \text{cm}^3}{1\, \text{L}} = 1500\, \text{cm}^3\] Now, we can find the mass of the water: \[m = \rho \times V = 0.997\, \frac{\text{g}}{\text{cm}^3} \times 1500 \, \text{cm}^3 \approx 1495.5\, \text{g}\] The mass of the water is approximately 1495.5 g.

Step 2: Calculate the volume of the ice

Now, we need to find the volume occupied by the ice by using the mass and the density of the ice at -10°C. We are given the density of the ice as 0.917 g/cm³. We can find the volume by rearranging the formula for finding mass: \[V_\text{ice} = \frac{m}{\rho_\text{ice}} = \frac{1495.5\, \text{g}}{0.917\, \frac{\text{g}}{\text{cm}^3}} \approx 1630.4\, \text{cm}^3\] The volume of the ice is approximately 1630.4 cm³.

Step 3: Can the ice be contained within the bottle?

Finally, we need to determine if the bottle can contain the ice. We do this by comparing the volume of the ice with the volume of the bottle. \[V_\text{ice} = 1630.4\, \text{cm}^3 > 1500\, \text{cm}^3 = V_\text{bottle}\] Since the volume of the ice is greater than the volume of the bottle, the ice cannot be contained within the bottle.

Key Concepts

Conversion of Units

Understanding unit conversion is key in solving problems involving physical properties. Here, we need to convert the volume of a bottle from liters to cubic centimeters (cm³), as calculations often require consistent units. To translate the given volume of 1.50 liters into cm³, you use the conversion factor between liters and cubic centimeters. Since 1 liter equals 1000 cm³, the calculation becomes:

• 1.50 L × 1000 cm³/L = 1500 cm³

This conversion allows us to use the density given in g/cm³ seamlessly in subsequent calculations.

Mass and Volume Calculation

Calculating mass and volume involves understanding the relationship described by the density formula: \(\rho = \frac{m}{V}\). Density (\(\rho\)) is mass per unit volume. To find mass using density and volume, rearrange the formula to: \(m = \rho \times V\).
For the water:

• Density = 0.997 g/cm³
• Volume = 1500 cm³
• Mass = 0.997 g/cm³ × 1500 cm³ = 1495.5 g

Now, using this mass, calculate the volume the ice occupies using the ice's density. Given:

• Ice density = 0.917 g/cm³
• Volume of ice, \(V_{\text{ice}} = \frac{1495.5\, \text{g}}{0.917\, \text{g/cm}^3} \approx 1630.4 \, \text{cm}^3\)

Density Comparison

When comparing densities, it's about understanding how different substances occupy space for the same mass. Water becomes ice, and due to its lower density (0.917 g/cm³ at -10°C compared to water at 0.997 g/cm³), the same mass expands into a larger volume.
This can cause practical issues, like whether the ice can be contained within the original volume. Comparing:

• Volume of bottle: 1500 cm³
• Volume of ice: 1630.4 cm³

The ice has a greater volume than the bottle's capacity, illustrating how density differences affect volume, despite having the same mass.