Question

Ethylene glycol (antifreeze) has a density of 1.11 \(\mathrm{g} / \mathrm{cm}^{3} .\) a. What is the mass in \(g\) of 417 \(\mathrm{mL}\) of this liquid? b. What is the volume in \(\mathrm{L}\) of 4.1 \(\mathrm{kg}\) of this liquid?

Short answer

a. The mass of 417 \(\mathrm{mL}\) of ethylene glycol is 462.87 \(\mathrm{g}\). b. The volume of 4.1 \(\mathrm{kg}\) of ethylene glycol is 3.6937 \(\mathrm{L}\).

Step-by-step solution

Calculate the mass of ethylene glycol

Use the density formula which is defined as mass divided by volume (density = mass/volume). Rearrange the formula to find mass (mass = density * volume). Density of ethylene glycol is given as 1.11 \(\mathrm{g}/\mathrm{cm}^3\), and the volume is 417 \(\mathrm{mL}\). Since 1 \(\mathrm{mL}\) is equivalent to 1 \(\mathrm{cm}^3\), we can directly multiply the density by the volume to get the mass (mass = 1.11 \(\mathrm{g}/\mathrm{cm}^3\) * 417 \(\mathrm{mL}\)).

Calculate the volume of ethylene glycol

Again, use the density formula but this time to find volume (volume = mass/density). Mass of ethylene glycol is given as 4.1 \(\mathrm{kg}\), which needs to be converted into grams since the density is given in \(\mathrm{g}/\mathrm{cm}^3\). We do this by multiplying the mass by 1,000 (since 1 \(\mathrm{kg}\) equals 1,000 \(\mathrm{g}\)). This gives us 4,100 \(\mathrm{g}\). Now, divide the mass in grams by the density to get the volume in \(\mathrm{cm}^3\) or \(\mathrm{mL}\) (volume = 4,100 \(\mathrm{g}\) / 1.11 \(\mathrm{g}/\mathrm{cm}^3\)). To convert the volume into liters, divide it by 1,000 (since 1 \(\mathrm{L}\) equals 1,000 \(\mathrm{mL}\)).

Key concepts

Mass-Volume Relationship

Understanding the mass-volume relationship is essential when dealing with substances in chemistry. This concept describes how the mass of a substance is related to its volume. Most know this relationship as density, which is the mass per unit volume of a substance. Density is widely used to identify substances and can be affected by temperature and pressure.

For example, the density of ethylene glycol (antifreeze) is provided as 1.11 g/cm³. To calculate the mass of a specific volume, you multiply the volume by the density. Conversely, to find the volume of a known mass, you divide the mass by the density, a process demonstrated in the exercise for ethylene glycol. Remember that the mass-volume relationship is a starting point for a host of important calculations in chemistry.

Unit Conversion

Unit conversion is a critical skill in chemistry as it allows for the comparison and computation of quantities in different units of measurement. In the exercise, we encounter two different units: grams for mass and milliliters (or cubic centimeters) for volume, which are straightforward to use together because 1 mL is equal to 1 cm³. However, when dealing with kilograms and liters, conversion becomes vital. To work with the provided density, we need to convert kilograms to grams (1 kg = 1000 g) and cubic centimeters to liters (1 L = 1000 cm³ or mL).

These conversions align the units with those used for density, enabling accurate calculations. It’s essential to keep track of units throughout a problem to ensure that the final answer is in the appropriate form, a practice clearly needed when we transitioned from grams to kilograms and milliliters to liters in the solution to the exercise.

Chemical Properties

Chemical properties describe a substance's behavior when it undergoes a chemical change or reaction, differentiating it from related physical properties which can be observed without changing the substance's structure. While density is a physical property, understanding the chemical properties of a substance like ethylene glycol is important for practical applications. Antifreeze, which contains ethylene glycol, has chemical properties that prevent water from freezing and overheating the engine, showcasing the relevance of such properties in the real world.

Chemical properties may include reactivity with acids or bases, flammability, or the ability to undergo specific chemical reactions. Comprehending these properties can inform how a substance might be used or how to handle it safely, demonstrating why a grounding in such characteristics is part of a comprehensive chemical education.