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Chapter 11: Problem 9

An antifreeze solution is made by mixing ethylene glycol \(\left(\rho=1116 \mathrm{~kg} / \mathrm{m}^{3}\right)\) with water. Suppose the specific gravity of such a solution is \(1.0730 .\) Assuming that the total volume of the solution is the sum of its parts, determine the volume percentage of ethylene glycol in the solution.

Short Answer

Expert verified
The volume percentage of ethylene glycol in the solution is approximately 62.93%.

Step by step solution

01

Understand the Problem

We need to find the volume percentage of ethylene glycol in an antifreeze solution where the specific gravity of the solution is 1.0730. We are informed that the specific gravity of ethylene glycol is derived from its density, which is 1116 kg/m\(^3\). The specific gravity of a substance is the ratio of its density compared to the density of water (1000 kg/m\(^3\)). Since the solution is a mix, we assume its total volume is the sum of individual component volumes (ethylene glycol and water).
02

Use the Formula for Specific Gravity

The specific gravity \( S \) of the solution is given by the formula:\[ S = \frac{\text{density of solution}}{\text{density of water}} \]We are given that \( S = 1.0730 \), hence:\[ \text{density of solution} = 1.0730 \times 1000 \text{ kg/m}^3 = 1073 \text{ kg/m}^3 \]
03

Set Up the Equation for Density

The density of the solution \( \rho_{solution} \) is determined by the mass and volume of its components:\[ \rho_{solution} = \frac{\text{mass of ethylene glycol} + \text{mass of water}}{\text{volume of ethylene glycol} + \text{volume of water}} \]Since the mass of each component can be expressed as a product of its density and volume, we write:\[ 1073 = \frac{1116V_{eg} + 1000V_{w}}{V_{eg} + V_{w}} \]where \( V_{eg} \) and \( V_{w} \) are the volumes of ethylene glycol and water, respectively.
04

Solve for Volume Ratio

We can equate from the formula:\[ 1073(V_{eg} + V_{w}) = 1116V_{eg} + 1000V_{w} \]Rearrange and solve for the ratio \( \frac{V_{eg}}{V_{w}} \):\[ 1073V_{eg} + 1073V_{w} = 1116V_{eg} + 1000V_{w} \]\[ 1073V_{w} - 1000V_{w} = 1116V_{eg} - 1073V_{eg} \]\[ 73V_{w} = 43V_{eg} \]Thus, \( \frac{V_{eg}}{V_{w}} = \frac{73}{43} \).
05

Calculate Volume Percentage of Ethylene Glycol

From the previous step, the proportion of ethylene glycol in the entire solution can be found:\[ V_{eg} = \frac{73}{43+73}(V_{eg} + V_{w}) = \frac{73}{116}(V_{eg} + V_{w}) \]This gives the volume percentage of ethylene glycol as:\[ \frac{V_{eg}}{V_{eg} + V_{w}} \times 100 = \frac{73}{116} \times 100 \approx 62.93\% \]

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Key Concepts

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

Density
Density is a measure of how much mass is contained within a certain volume. In scientific terms, it is defined as the mass per unit volume. The formula to calculate density is given as:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]Understanding density is crucial as it determines how substances interact when mixed together. For example, substances with different densities may separate over time because the denser component will sink lower than the less dense component.
In the context of our exercise, density plays a key role in calculating the specific gravity and subsequently finding out how the volumes of ethylene glycol and water contribute to the overall solution's properties.
Solution Mixing
Solution mixing involves the combination of two or more substances; in this case, we are mixing ethylene glycol and water to create an antifreeze solution. When two substances are mixed:
  • Their individual densities and volumes must be considered to predict the density of the resulting solution.
  • It is often assumed that the total volume of the mixture equals the sum of the individual volumes of the ingredients, which assumes no volume change upon mixing.
For antifreeze solutions, achieving a certain density is crucial to ensure it performs safely under various temperature conditions. Properly mixed solutions prevent the engine's coolant from freezing or boiling over, offering essential protection throughout various climates.
Volume Percentage
Volume percentage is a way of expressing concentrations in solutions and helps to determine how much of each component is present in a mixture by volume. To calculate the volume percentage of ethylene glycol in the antifreeze solution, the formula is:
\[ \text{Volume Percentage} = \frac{\text{Volume of ethylene glycol}}{\text{Total Volume of the solution}} \times 100 \]In the solution, knowing the volume percentage helps determine the effectiveness of the antifreeze. This ensures that the concentration of ethylene glycol is adequate to provide the expected antifreeze properties without being wasteful or inefficient. Calculating volume percentage is critical to tailor the solution to specific temperature ranges and performance requirements.
Ethylene Glycol
Ethylene glycol is an organic compound commonly used as an antifreeze and in de-icing solutions. It has a high density of 1116 kg/m\(^3\), which influences its behavior in a mixed solution. Here are some important aspects to consider:
  • Due to its higher density compared to water (1000 kg/m\(^3\)), solutions mixed with ethylene glycol are typically denser.
  • It is a crucial component for preventing freezing in automotive and other industrial applications.
  • The specific gravity, being 1.0730 in our case, means the solution is denser than water, which impacts its ability to prevent freezing at lower temperatures.
Understanding ethylene glycol's role in solution helps us appreciate the importance of its concentration in ensuring the antifreeze solution performs efficiently. Adjusting the percentage of ethylene glycol ultimately determines how effectively a solution can operate under varying temperature conditions.

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Most popular questions from this chapter

A 1.3-m length of horizontal pipe has a radius of \(6.4 \times 10^{-3} \mathrm{~m}\). Water within the pipe flows with a volume flow rate of \(9.0 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{s}\) out of the right end of the pipe and into the air. What is the pressure in the flowing water at the left end of the pipe if the water behaves as (a) an ideal fluid and (b) a viscous fluid \(\left(\eta=1.00 \times 10^{-3} \mathrm{~Pa} \cdot \mathrm{s}\right)\) ?

Two hoses are connected to the same outlet using a Y- connector, as the drawing shows. The hoses \(A\) and \(B\) have the same length, but hose \(B\) has the larger radius. Each is open to the atmosphere at the end where the water exits. Water flows through both hoses as a viscous fluid, and Poiseuille's law \({ }^{[} Q=\pi R^{4}\left(P_{2}-P_{1}\right) /(8 \eta L)^{]}\) applies to each. In this law, \(P_{2}\) is the pressure upstream, \(P_{1}\) is the pressure downstream, and \(Q\) is the volume flow rate. (a) For hoses \(\mathrm{A}\) and \(\mathrm{B}\), is the value for the term \(P_{2}-P_{1}\) the same or different? (b) How is \(Q\) related to the radius of a hose and the speed of the water in the hose? Account for your answers.

In a very large closed tank, the absolute pressure of the air above the water is \(6.01 \times 10^{5} \mathrm{~Pa}\). The water leaves the bottom of the tank through a nozzle that is directed straight upward. The opening of the nozzle is \(4.00 \mathrm{~m}\) below the surface of the water. (a) Find the speed at which the water leaves the nozzle. (b) Ignoring air resistance and viscous effects, determine the height to which the water rises.

One of the concrete pillars that support a house is \(2.2 \mathrm{~m}\) tall and has a radius of \(0.50 \mathrm{~m}\). The density of concrete is about \(2.2 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). Find the weight of this pillar in pounds \((1 \mathrm{~N}=0.2248 \mathrm{lb})\).

A gold prospector finds a solid rock that is composed solely of quartz and gold. (a) How is the total mass \(m_{\mathrm{T}}\) of the rock related to the mass \(m_{\mathrm{G}}\) of the gold and the mass \(m_{\mathrm{Q}}\) of the quartz? (b) What is the relationship between the total volume \(V_{\mathrm{T}}\) of the rock, the volume \(V_{\mathrm{G}}\) of the gold, and the volume \(V_{\mathrm{Q}}\) of the quartz? (c) How is the volume of a substance (gold or quartz) related to the mass of the substance and its density? The mass and volume of the rock are, respectively, \(12.0 \mathrm{~kg}\) and \(4.00 \times 10^{-3} \mathrm{~m}^{3}\). Find the mass of the gold in the rock.
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