Problem 155 - General Chemistry

Question

A \(47.0-\mathrm{cm}^{3}\) sample of ocean water has a density of \(1025 \mathrm{~kg} / \mathrm{m}^{3}\). Assuming all of the solids stay in solution, what will be the density of this water sample in units of \(\mathrm{g} / \mathrm{cm}^{3}\) after \(4.1 \mathrm{~mL}\) of water has evaporated?

Step-by-Step Solution

Answer

The new density of the seawater sample is approximately 1.122 g/cm³.

Step 1: Calculate initial mass of seawater

We need to find the initial mass of the 47.0 cm³ seawater sample. Use the formula for mass: \( \text{mass} = \text{density} \times \text{volume} \). The density is given as 1025 kg/m³. Convert the unit of density to g/cm³ by using 1 kg/m³ = 0.001 g/cm³. Thus, \( 1025 \text{ kg/m}^3 = 1.025 \text{ g/cm}^3 \). Calculate initial mass: \( 1.025 \text{ g/cm}^3 \times 47.0 \text{ cm}^3 = 48.175 \text{ g} \).

Step 2: Calculate final volume after evaporation

The problem states that 4.1 mL of water has evaporated. Since 1 mL = 1 cm³, the volume of water left is \( 47.0 \text{ cm}^3 - 4.1 \text{ cm}^3 = 42.9 \text{ cm}^3 \).

Step 3: Calculate final density of seawater

Now we determine the new density of the seawater sample by using the same mass but reduced volume. Use the formula \( \text{density} = \frac{\text{mass}}{\text{volume}} \). With the mass remaining as 48.175 g and the new volume as 42.9 cm³, the final density is \( \frac{48.175 \text{ g}}{42.9 \text{ cm}^3} \approx 1.122 \text{ g/cm}^3 \).

Key Concepts

Ocean Water Density

Ocean water density plays a crucial role in understanding the characteristics of seawater. In general, density is a measure of how much mass is contained in a given volume. For water, especially ocean water, density depends on several factors such as temperature, salinity, and pressure. Ocean water typically has a higher density than fresh water due to its salt content.

In the given problem, the density of ocean water is specified as 1025 kg/m³. This value indicates that each cubic meter of ocean water has a mass of 1025 kilograms. High salinity increases density because the dissolved salts add extra mass without significantly increasing the volume. Understanding the adjustments in mass and volume can help us figure out the water's density after certain changes, like evaporation.

Evaporation Effect on Density

Evaporation is the process where water molecules escape from a liquid into a vapor phase. When water evaporates from ocean water, the volume decreases, but salts and other solutes remain behind in the solution. This results in a higher concentration of salts, and therefore, an increase in density.

In the exercise, 4.1 mL of ocean water evaporates. Since 1 mL equals 1 cm³, the remaining water volume reduces to 42.9 cm³ from the original 47.0 cm³. Despite the volume loss, the mass remains constant at 48.175 g because evaporation only removes water, not the dissolved solids. As a result, the density of the ocean water increases from its initial value. We calculate the new density by dividing the mass by the new volume, leading to a higher density of approximately 1.122 g/cm³.

Metric Unit Conversion

Metric unit conversion is a vital skill in science, allowing us to understand measurements across different scales. Density can be expressed in various units such as kg/m³ or g/cm³, depending on the context or region.

In this exercise, the ocean water's density is initially provided in kg/m³. To perform calculations compatible with the volume of seawater measured in cm³, it's necessary to convert this density to g/cm³. This involves understanding the relationship between these units: 1 kg equals 1000 grams and 1 m³ equals 1,000,000 cm³. From this, we know that 1 kg/m³ equals 0.001 g/cm³. By converting 1025 kg/m³ to 1.025 g/cm³, calculations become more straightforward, matching the volume unit, and allowing precise determination of the mass and post-evaporation density.