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Which liquid will occupy the greater volume, \(50 \mathrm{~g}\) of water or \(50 \mathrm{~g}\) of ethyl alcohol? Explain.

Short Answer

Expert verified
Ethyl alcohol occupies a greater volume of approximately 63.38 cm³ compared to water's 50 cm³.

Step by step solution

01

Understanding the Concept

The volume of a liquid is calculated using its density and mass. The formula to find the volume is: \[ V = \frac{m}{\rho} \]Where \(V\) is the volume, \(m\) is the mass, and \(\rho\) is the density.
02

Identify the Densities

Find the densities of water and ethyl alcohol. The density of water is approximately \(1 \mathrm{~g/cm^3}\) and the density of ethyl alcohol is approximately \(0.789 \mathrm{~g/cm^3}\).
03

Calculate the Volume of Water

Using the formula \[ V = \frac{m}{\rho} \]Substitute the mass of water (50 g) and the density of water (1 g/cm³): \[ V_{\text{water}} = \frac{50 \mathrm{~g}}{1 \mathrm{~g/cm^3}} = 50 \mathrm{~cm^3} \]
04

Calculate the Volume of Ethyl Alcohol

Using the same formula \[ V = \frac{m}{\rho} \]Substitute the mass of ethyl alcohol (50 g) and the density of ethyl alcohol (0.789 g/cm³): \[ V_{\text{ethyl~alcohol}} = \frac{50 \mathrm{~g}}{0.789 \mathrm{~g/cm^3}} \approx 63.38 \mathrm{~cm^3} \]
05

Compare the Volumes

Compare the volumes of water and ethyl alcohol from the previous steps: \[ V_{\text{water}} = 50 \mathrm{~cm^3} \]\[ V_{\text{ethyl~alcohol}} \approx 63.38 \mathrm{~cm^3} \]Since 63.38 cm³ is greater than 50 cm³, the ethyl alcohol occupies a greater volume.

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Key Concepts

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

Concept of Density
Density is a measure of how much mass is contained in a given volume. It’s like trying to figure out how tightly packed the particles are within a substance. The formula for density is simple: \( \rho = \frac{m}{V} \), where \( \rho \) represents density, \( m \) is mass, and \( V \) is volume.
For instance, water has a density of about \(1 \text{ g/cm}^3 \), meaning 1 gram of water fits into 1 cubic centimeter. Densities can vary vastly among different materials; solids are generally denser than liquids, and liquids are denser than gases. Understanding density is crucial because it helps in identifying substances and is fundamental in calculations involving mass and volume.
For example, if you have 50 grams of water (\text{density] = 1 \text{ g/cm}^3) and 50 grams of ethyl alcohol (\text{density} \approx 0.789 \text{ g/cm}^3), we use their densities to find out how much volume each will occupy.
Understanding Volume
Volume measures the amount of space an object or substance occupies. Imagine a container: the volume is the space inside that container. You can find the volume of a liquid by using the formula \( V = \frac{m}{\rho} \), where \( V \) is volume, \( m \) is mass, and \( \rho \) is the density.
For water with a mass of 50 grams and density of 1 \text{ g/cm}^3, the volume is \( \frac{50 \text{ g}}{1 \text{ g/cm}^3} = 50 \text{ cm}^3 \).
On the other hand, for ethyl alcohol with a mass of 50 grams and a lower density of about 0.789 \text{ g/cm}^3, the volume is \( \frac{50 \text{ g}}{0.789 \text{ g/cm}^3} \approx 63.38 \text{ cm}^3 \).
Here, the lower density of ethyl alcohol means it will occupy more space than water for the same mass.
Explaining Mass
Mass is a measure of the amount of matter in an object, usually measured in grams (g), kilograms (kg), etc. It differs from weight, which is the force exerted by gravity on that mass. In calculations involving density and volume, mass is a key component and is generally provided or can be measured easily.
In our exercise, we have a mass of 50 grams for both water and ethyl alcohol. This common mass allows us to compare how differently they behave due to their different densities.
With a known density and mass, we can straightforwardly determine the volume that each substance will occupy. Understanding the relations between mass, density, and volume is crucial for solving such problems effectively, providing insights into the material properties and their applications.

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Most popular questions from this chapter

Solve the following problems, stating answers to the proper number of significant figures: (a) \(12.62+1.5+0.25=\) ? (b) \(\left(2.25 \times 10^{3}\right)\left(4.80 \times 10^{4}\right)=\) ? (c) \(\frac{(452)(6.2)}{14.3}=\) ? (d) \((0.0394)(12.8)=\) ? (e) \(\frac{0.4278}{59.6}=\) ? (f) \(10.4+3.75\left(1.5 \times 10^{4}\right)=\) ?

Complete the following metric conversions using the correct number of significant figures: (a) \(4.5 \mathrm{~cm}\) to \(\hat{\mathrm{A}}\) (b) \(12 \mathrm{~nm}\) to \(\mathrm{cm}\) (c) \(8.0 \mathrm{~km}\) to \(\mathrm{mm}\) (d) \(164 \mathrm{mg}\) to \(\mathrm{g}\) (e) \(0.65 \mathrm{~kg}\) to \(\mathrm{mg}\) (f) \(5.5 \mathrm{~kg}\) to g (g) \(0.468 \mathrm{~L}\) to \(\mathrm{mL}\) (h) \(9.0 \mu \mathrm{L}\) to \(\mathrm{mL}\)

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A Mexican restaurant has the reputation of serving the hottest salsa in town. In order to maintain this standard, the restaurant must use a minimum of \(8.41 \%\) by mass of cayenne pepper. If the restaurant just received a \(3.26-\mathrm{kg}\) shipment of cayenne pepper, how many pounds of salsa can it make?

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