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Find the density and specific gravity of ethyl alcohol if \(63.3 \mathrm{~g}\) occupies \(80.0 \mathrm{~mL}\).

Short Answer

Expert verified
Density is 0.791 g/mL; Specific gravity is 0.791.

Step by step solution

01

Calculate the Density

Density is defined as mass per unit volume. To calculate the density, use the formula \( \rho = \frac{m}{V} \), where \( m \) is mass and \( V \) is volume. Here, \( m = 63.3 \text{ g} \) and \( V = 80.0 \text{ mL} \).Substitute these values into the formula:\[ \rho = \frac{63.3 \text{ g}}{80.0 \text{ mL}} = 0.79125 \text{ g/mL} \]
02

Calculate the Specific Gravity

Specific gravity is the ratio of the density of a substance to the density of water. Since the density of water is approximately \( 1 \text{ g/mL} \) at room temperature, use the formula for specific gravity:\[ \text{Specific Gravity} = \frac{\text{Density of Ethyl Alcohol}}{\text{Density of Water}} = \frac{0.79125 \text{ g/mL}}{1 \text{ g/mL}} = 0.79125 \]

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Key Concepts

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

Specific Gravity
Specific gravity is an important concept in understanding how dense a substance is compared to water. Essentially, it is a ratio, meaning it compares two similar quantities. To calculate specific gravity, you divide the density of the substance by the density of water.

Since the density of water is typically 1 g/mL (at room temperature), calculating specific gravity becomes straightforward. For ethyl alcohol, if the density is calculated at 0.79125 g/mL, its specific gravity is \[\text{Specific Gravity} = \frac{0.79125 \text{ g/mL}}{1 \text{ g/mL}} = 0.79125\]

This comparison reveals that ethyl alcohol is less dense than water, as its specific gravity is less than 1. This property is useful in applications where alcohol needs to float or diffuse differently than water.
Density Formula
The density formula is crucial for calculating how much mass is present in a specific volume of a material. The formula is represented as \[\rho = \frac{m}{V}\]where:
  • \(\rho\) represents density
  • \(m\) is the mass of the substance
  • \(V\) is the volume the mass occupies
To calculate density, simply divide the mass by the volume. If you have a mass of 63.3 grams of ethyl alcohol occupying 80.0 mL, you substitute these values into the formula:\[\rho = \frac{63.3 \text{ g}}{80.0 \text{ mL}} = 0.79125 \text{ g/mL}\]

This calculation helps anyone determine how tightly packed the molecules are in a given space, which is crucial for material identification and usage in various fields.
Ethyl Alcohol Properties
Ethyl alcohol, also known as ethanol, holds specific properties that make it useful in various applications, from manufacturing to medical use. Ethanol's density is lower than that of water, making it lighter per unit of volume. This property is represented through its density value of approximately 0.79 g/mL.

Key properties of ethyl alcohol include:
  • It is volatile, which means it evaporates quickly at room temperature.
  • Ethanol is a good solvent, meaning it can dissolve various substances effectively.
  • It is flammable and can burn easily, which is important for fuel applications.
These properties contribute to its widespread use not only in alcoholic beverages but also in cleaning solutions, fuels, and even biofuels.
Mass and Volume Relationship
Understanding the relationship between mass and volume is foundational in science. Mass refers to the quantity of matter contained in a substance, while volume measures the amount of space it takes up. When it comes to calculating density, you need to know how mass and volume correspond. The formula for density \[\rho = \frac{m}{V}\]clearly shows this relationship.

For example, if you increase the mass while keeping the volume constant, the density will rise, indicating a more "packed" substance. Conversely, increasing the volume with a fixed mass will reduce density, reflecting a more "spread out" substance.

This concept helps us assess and compare substances based on how much mass they have in a given space, illustrating differences in material properties and uses.

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Most popular questions from this chapter

Find the density and specific gravity of gasoline if 51 g occupies 75 \(\mathrm{cm}^{3}\). Make sure you know how to convert cubic centimeters to cubic meters: \(1.0 \mathrm{~m}^{3}=1.0 \times 10^{6} \mathrm{~cm}^{3}\). 1 $$ \begin{array}{l} \text { Density }=\frac{\text { Mass }}{\text { Volume }}=\frac{0.051 \mathrm{~kg}}{75 \times 10^{-6} \mathrm{~m}^{3}}=6.8 \times 10^{2} \mathrm{~kg} / \mathrm{m}^{3} \\ \text { sp gr }=\frac{\text { Density of gasoline }}{\text { Density of water }}=\frac{6.8 \times 10^{2} \mathrm{~kg} / \mathrm{m}^{3}}{1000 \mathrm{~kg} / \mathrm{m}^{3}}=0.68 \\ \text { sp gr }=\frac{\text { Mass of } 75 \mathrm{~cm}^{3} \text { gasoline }}{\text { Mass of } 75 \mathrm{~cm}^{3} \text { water }}=\frac{51 \mathrm{~g}}{75 \mathrm{~g}}=0.68 \end{array} $$

A vertical wire \(5.0 \mathrm{~m}\) long and of \(0.008 \mathrm{~cm}^{2}\) cross- sectional area has a modulus \(Y=200\) GPa. A 2.0-kg object is fastened to its end and stretches the wire elastically. If the object is now pulled down a little and released, the object undergoes vertical SHM. Find the period of its vibration. The force constant of the wire acting as a vertical spring is given by \(k=F / \Delta L\), where \(\Delta L\) is the deformation produced by the force (weight) \(F\). But, from F/A \(=\mathrm{Y}\left(\Delta \mathrm{L} / \mathrm{L}_{0}\right)\), $$ k=\frac{F}{\Delta L}=\frac{A Y}{L_{0}}=\frac{\left(8.8 \times 10^{-7} \mathrm{~m}^{2}\right)\left(2.00 \times 10^{11} \mathrm{~Pa}\right)}{5.0 \mathrm{~m}}=35 \mathrm{kN} / \mathrm{m} $$ Then for the period we have $$ T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{2.0 \mathrm{~kg}}{35 \times 10^{3} \mathrm{~N} / \mathrm{m}}}=0.047 \mathrm{~s} $$

A 60 -kg woman stands on a light, cubical box that is \(5.0 \mathrm{~cm}\) on each edge. The box sits on the floor. What pressure does the box exert on the floor? $$ P=\frac{F}{A}=\frac{(60)(9.81) \mathrm{N}}{\left(5.0 \times 10^{-2} \mathrm{~m}\right)^{2}}=2.4 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2} $$

Atmospheric pressure is about \(1.01 \times 10^{5} \mathrm{~Pa} .\) How large a force does the atmosphere exert on a \(2.0-\mathrm{cm}^{2}\) area on the top of your head? Because \(P=F / A\), where \(F\) is perpendicular to \(A\), we have \(F=P A\). Assuming that \(2.0 \mathrm{~cm}^{2}\) of your head is flat (nearly correct) and that the force due to the atmosphere is perpendicular to the surface (as it is), $$ F=P A=\left(1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(2.0 \times 10^{-4} \mathrm{~m}^{2}\right)=20 \mathrm{~N} $$

Determine the volume of \(200 \mathrm{~g}\) of carbon tetrachloride, for which sp gr \(=1.60\).

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