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At \(25^{\circ} \mathrm{C}\) the density of water is \(0.997 \mathrm{g} / \mathrm{cm}^{3},\) whereas the density of ice at \(-10^{\circ} \mathrm{C}\) is \(0.917 \mathrm{g} / \mathrm{cm}^{3} .\) (a) If a soft-drink can (volume \(=250 . \mathrm{mL}\) ) is filled completely with pure water at \(25^{\circ} \mathrm{C}\) and then frozen at \(-10^{\circ} \mathrm{C},\) what volume does the solid occupy? (b) Can the ice be contained within the can?

Short Answer

Expert verified
The ice cannot be contained within the can as it expands to approximately 271.91 cm³.

Step by step solution

01

Convert Water Mass to Grams

First, convert the volume of the water in the can into grams using the density of water at 25°C. Since the density of water is 0.997 g/cm³ and 1 mL is equivalent to 1 cm³, multiply the volume by the density:\[\text{mass of water} = 250 \, \text{mL} \times 0.997 \, \text{g/mL} = 249.25 \, \text{g}.\]
02

Calculate Ice Volume

Now calculate the volume that the ice will occupy. Use the density of ice at -10°C, which is 0.917 g/cm³, to find the volume of 249.25 g of ice:\[\text{volume of ice} = \frac{249.25 \, \text{g}}{0.917 \, \text{g/cm}^3} \approx 271.91 \, \text{cm}^3.\]
03

Compare Ice Volume with Can Volume

Compare the volume the ice occupies with the volume of the soft-drink can. The volume of the can is 250 cm³, and the volume of ice is approximately 271.91 cm³. Since 271.91 cm³ > 250 cm³, the ice occupies more volume than the can can handle.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Water Density
Water density is a crucial concept when understanding how much space water takes up in different conditions. Density is defined as mass per unit volume, and for water, it depends on the temperature. At 25°C, the density of water is approximately 0.997 grams per cubic centimeter (g/cm³). This means that one milliliter (mL) of water weighs about 0.997 grams.

This figure is important because it allows us to calculate the mass of water when the volume is known, or vice versa. For instance, if you have a 250 mL soft drink can filled with water at 25°C, you can calculate its mass by multiplying the volume (250 mL) by the density (0.997 g/cm³), resulting in 249.25 grams. This conversion from volume to weight helps in further calculations, such as when determining the volume of ice formed from the water when it's frozen.
Ice Density
Ice density is slightly different from water density because the structure of ice causes it to take up more space. At -10°C, ice has a density of 0.917 g/cm³. This lower density compared to liquid water is why ice floats.

When water freezes, the molecules form a crystalline structure that is more open than that of liquid water, causing the ice to expand. To find out how much space ice takes, we use its density. For example, if the mass of the ice is 249.25 grams (as calculated from the mass of the water before freezing), the volume can be determined by dividing the mass by the ice's density: 249.25 g / 0.917 g/cm³, resulting in approximately 271.91 cm³. This increase in volume is crucial to consider when containing ice within a defined space.
Volume Calculation
Volume calculation is a fundamental step when dealing with substances that change state, like water turning into ice. It involves converting mass into volume using density, helping us to understand the spatial requirements of different substances.

In the exercise, we start by calculating the mass of the water, which is already in a volume of 250 mL at 25°C. By multiplying this volume by the density of water, we get the water's mass. Once the water is frozen, we calculate the space the ice will occupy. By using the mass of water and the density of ice, we determine that the ice will have a volume of approximately 271.91 cm³.
  • This step is crucial because it allows you to compare the volume of the ice with the original volume of the container, which was 250 cm³.
  • Knowing whether the ice can fit within the container is vital for practical purposes and prevention of overflow or container damage.
  • Effective volume calculation helps in anticipating such scenarios, especially in real-world applications and experiments.
By understanding these calculations, we can practically manage situations involving phase changes of water and other substances.

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Most popular questions from this chapter

Solve the following equation for the unknown value, \(C\). $$ (0.502)(123)=(750 .) C $$

Express the following numbers in fixed notation (e.g., \(\left.123 \times 10^{2}=123\right)\) (a) \(1.62 \times 10^{3}\) (b) \(2.57 \times 10^{-4}\) (c) \(6.32 \times 10^{-2}\)

Suppose your bedroom is \(18 \mathrm{ft}\) long, \(15 \mathrm{ft}\) wide, and the distance from floor to ceiling is \(8 \mathrm{ft}, 6\) in. You need to know the volume of the room in metric units for some scientific calculations. (a) What is the room's volume in cubic meters? In liters? (b) What is the mass of air in the room in kilograms? In pounds? (Assume the density of air is \(1.2 \mathrm{g} / \mathrm{L}\) and that the room is empty of furniture.)

A The density of a single, small crystal can be determined by the flotation method. This method is based on the idea that if a crystal and a liquid have precisely the same density, the crystal will hang suspended in the liquid. A crystal that is more dense will sink; one that is less dense will float. If the crystal neither sinks nor floats, then the density of the crystal equals the density of the liquid. Generally, mixtures of liquids are used to get the proper density. Chlorocarbons and bromocarbons (see the list below) are often the liquids of choice. If the two liquids are similar, then volumes are usually additive and the density of the mixture relates directly to composition. (An example: \(1.0 \mathrm{mL}\) of \(\mathrm{CHCl}_{3}, d=1.4832 \mathrm{g} / \mathrm{mL},\) and 1.0 mL of \(\mathrm{CCl}_{4}, d=1.5940 \mathrm{g} / \mathrm{mL},\) when mixed, give \(2.0 \mathrm{mL}\) of a mixture with a density of \(1.5386 \mathrm{g} / \mathrm{mL} .\) The density of the mixture is the average of the values of the two individual components.) The problem: A small crystal of silicon, germanium, tin, or lead (Group 4A in the periodic table) will hang suspended in a mixture made of \(61.18 \%\) (by volume) \(\mathrm{CH}\) IBr \(_{3}\) and \(38.82 \%\) (by volume) \(\mathrm{CHCl}_{3} .\) Calculate the density and identify the element. (You will have to look up the values of the density of the elements in a manual such as the The Handbook of Chemistry and Physics in the library or in a World Wide Web site such as WebElements at, www.webelements.com.) $$\begin{array}{llll} \hline \text { Liquid } & \text { Density }(\mathrm{g} / \mathrm{mL}) & \text { Liquid } & \text { Density }(\mathrm{g} / \mathrm{mL}) \\ \hline \mathrm{CH}_{2} \mathrm{Cl}_{2} & 1.3266 & \mathrm{CH}_{2} \mathrm{Br}_{2} & 2.4970 \\ \mathrm{CH} \mathrm{Cl}_{3} & 1.4832 & \mathrm{CHBr}_{3} & 2.8899 \\ \mathrm{CCl}_{4} & 1.5940 & \mathrm{CBr}_{4} & 2.9609 \\ \hline \end{array}$$

In each case, decide whether the change is a chemical or physical change. (a) A cup of household bleach changes the color of your favorite T-shirt from purple to pink. (b) Water vapor in your exhaled breath condenses in the air on a cold day. (c) Plants use carbon dioxide from the air to make sugar. (d) Butter melts when placed in the sun.

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