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(a) Do you expect the viscosity of glycerol, \(\mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{OH})_{3}\) , to be larger or smaller than that of 1 -propanol, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\) ? (b) Explain. [ Section 11.3\(]\)

Short Answer

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(a) The viscosity of glycerol, \(\mathrm{C}_{3}\mathrm{H}_{5}(\mathrm{OH})_{3}\), is expected to be larger than that of 1-propanol, \(\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{OH}\). (b) This is because glycerol has more hydroxyl groups, allowing for more hydrogen bonding between its molecules, which creates greater resistance to flow and thus higher viscosity compared to 1-propanol that has fewer hydrogen bonds between its molecules.

Step by step solution

01

Understand the molecular structure of glycerol and 1-propanol

First, let's examine the molecular structures of glycerol and 1-propanol. Glycerol has the formula \(\mathrm{C}_{3}\mathrm{H}_{5}(\mathrm{OH})_{3}\), which means it contains 3 carbon atoms, 5 hydrogen atoms, and 3 hydroxyl (OH) groups. 1-Propanol has the formula \(\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{OH}\), which indicates that it has 3 carbon atoms, 7 hydrogen atoms, and 1 hydroxyl group.
02

Analyze the intermolecular forces

Now, we need to analyze the intermolecular forces between the molecules of the two substances. Intermolecular forces play a critical role in determining the viscosity of a substance. There are three main types of intermolecular forces to consider: London dispersion forces, dipole-dipole forces, and hydrogen bonding.
03

Compare the hydrogen bonding in glycerol and 1-propanol

Both glycerol and 1-propanol have polar functional groups (hydroxyl groups), which allow for hydrogen bonding between the molecules. However, glycerol has 3 hydroxyl groups while 1-propanol has only 1. This means that each glycerol molecule can participate in more hydrogen bonding compared to 1-propanol. As hydrogen bonding is the strongest type of intermolecular force, this factor will significantly impact the relative viscosity of the two substances.
04

Compare the London dispersion forces and dipole-dipole forces

The London dispersion forces and dipole-dipole forces should also be considered. However, due to similar molecular size and polarities of glycerol and 1-propanol, these forces will not have a significant difference between the two substances. Therefore, the hydrogen bonding will be the dominant factor in comparing the viscosities.
05

Predict and explain the relative viscosities of glycerol and 1-propanol

(a) Given that glycerol can participate in more hydrogen bonding than 1-propanol due to its greater number of hydroxyl groups, we can predict that the viscosity of glycerol will be larger than that of 1-propanol. (b) The explanation for this is that the increased hydrogen bonding in glycerol creates more resistance to flow, leading to a higher viscosity compared to 1-propanol which has fewer hydrogen bonds between its molecules.

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Key Concepts

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

Molecular Structure
Understanding the molecular structure of a compound is crucial in predicting its properties, such as viscosity. The molecular structure describes the arrangement of atoms within a molecule and can greatly influence how the molecule interacts with others.
In the case of glycerol ( C_{3}H_{5}(OH)_{3} ) and 1-propanol ( C_{3}H_{7}OH ), both molecules share a backbone of three carbon atoms. However, glycerol has three hydroxyl ( OH ) groups whereas 1-propanol contains just one. This distinction is key.
  • Glycerol's additional hydroxyl groups make its molecular structure more complex.
  • The extra OH groups also increase the potential for interactions between glycerol molecules through hydrogen bonding.
The more intricate the molecular structure, typically, the more potent the effects of intermolecular interactions like hydrogen bonds will be.
Intermolecular Forces
Intermolecular forces are interactions between molecules, which can affect a substance's physical properties. There are three primary types of intermolecular forces: London dispersion forces, dipole-dipole forces, and hydrogen bonding.
London dispersion forces are weak attractions present in all molecules, arising from temporary dipoles. They're significant in nonpolar molecules but less impactful here due to the polar nature of our molecules.
Dipole-dipole forces occur between polar molecules. Both glycerol and 1-propanol are polar due to their hydroxyl groups, enabling dipole-dipole interactions.
Finally, hydrogen bonding is a particularly strong form of dipole-dipole interaction, given the presence of OH groups. In this guide, we'll see that glycerol's increased ability to form hydrogen bonds plays a crucial role in determining its greater viscosity compared to 1-propanol.
Hydrogen Bonding
Hydrogen bonding is an impressive phenomenon where a hydrogen atom, covalently bonded to a highly electronegative atom like oxygen, interacts with another electronegative atom. This bond is relatively strong compared to other types of intermolecular forces and significantly affects compounds' physical properties.
In glycerol, the presence of three hydroxyl groups compared to the single hydroxyl group in 1-propanol allows for more hydrogen bond formation between molecules. This abundance of hydrogen bonding leads to greater intermolecular attraction in glycerol, thereby increasing its resistance to flow and, consequently, its viscosity.
  • More hydrogen bonds mean stronger intermolecular forces.
  • Stronger intermolecular forces result in higher viscosity because molecules tend to "stick" together more.
Therefore, glycerol's increased hydrogen bonding capability is directly responsible for its higher viscosity when compared to 1-propanol.

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Most popular questions from this chapter

For many years drinking water has been cooled in hot climates by evaporating it from the surfaces of canvas bags or porous clay pots. How many grams of water can be cooled from 35 to \(20^{\circ} \mathrm{C}\) by the evaporation of 60 \(\mathrm{g}\) of water?(The heat of vaporization of water in this temperature range is 2.4 \(\mathrm{kJ} / \mathrm{g} .\) The specific heat of water is \(4.18 \mathrm{J} / \mathrm{g}-\mathrm{K}\) .

Appendix \(\mathrm{B}\) lists the vapor pressure of water at various external pressures. (a) Plot the data in Appendix B, vaporpressure (torr) versus temperature \(\left(^{\circ} \mathrm{C}\right) .\) From your plot, estimate the vapor pressure of water at body temperature, \(37^{\circ} \mathrm{C}\) (b) Explain the significance of the data point at 760.0 torr, \(100^{\circ} \mathrm{C}\) (c) A city at an altitude of 5000 \(\mathrm{ft}\)above sea level has a barometric pressure of 633 torr. To what temperature would you have to heat water to boil it in this city? (d) A city at an altitude of 500 ft below sea level would have a barometric pressure of 774 torr. To what temperature would you have to heat water to boil it in this city?

At \(25^{\circ} \mathrm{C}\) gallium is a solid with a density of 5.91 \(\mathrm{g} / \mathrm{cm}^{3} .\) Its melting point, \(29.8^{\circ} \mathrm{C},\) is low enough that you can melt it by holding it in your hand. The density of liquid gallium just above the melting point is 6.1 \(\mathrm{g} / \mathrm{cm}^{3} .\) Based on this information, what unusual feature would you expect to find in the phase diagram of gallium?

Which of the following affects the vapor pressure of a liquid? (a) Volume of the liquid, (b) surface area, (c) intermolecular attractive forces, (d) temperature, (e) density of the liquid.

(a) What is the relationship between surface tension and temperature? (b) What is the relationship between viscosity and temperature? (c) Why do substances with high surface tension also tend to have high viscosities?

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