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Suppose you have two colorless molecular liquids, one boiling at \(-84^{\circ} \mathrm{C},\) the other at \(34^{\circ} \mathrm{C},\) and both at atmospheric pressure. Which of the following statements is correct? For each statement that is not correct, modify the statement so that it is correct. (a) The higher-boiling liquid has greater total intermolecular forces than the lower- boiling liquid. (b) The lower-boiling liquid must consist of nonpolar molecules. (c) The lower- boiling liquid has a lower molecular weight than the higher-boiling liquid. (d) The two liquids have identical vapor pressures at their normal boiling points. (e) At \(-84^{\circ}\) both liquids have vapor pressures of 760 \(\mathrm{mm} \mathrm{Hg}\) .

Short Answer

Expert verified
The correct and modified statements are as follows: (a) The higher-boiling liquid has greater total intermolecular forces than the lower-boiling liquid. (Correct) (b) The lower-boiling liquid could consist of nonpolar molecules or have weaker intermolecular forces. (Modified) (c) The lower-boiling liquid may have a lower molecular weight or weaker intermolecular forces than the higher-boiling liquid. (Modified) (d) The two liquids have identical vapor pressures at their normal boiling points. (Correct) (e) At \( -84^{\circ}\), the liquid with a boiling point of \( -84^{\circ}\) has a vapor pressure of 760 \(\mathrm{mm} \mathrm{Hg}\), while the other liquid has a lower vapor pressure. (Modified)

Step by step solution

01

Statement (a) Analysis and verification

Statement (a): The higher-boiling liquid has greater total intermolecular forces than the lower-boiling liquid. This statement is correct. When the boiling point of a liquid is high, this suggests that the intermolecular forces holding the molecules together are stronger. It requires more energy (higher temperature) to break these intermolecular forces and cause the liquid to boil.
02

Statement (b) Analysis and verification

Statement (b): The lower-boiling liquid must consist of nonpolar molecules. This statement is not entirely correct. While it is true that nonpolar molecules generally have weaker intermolecular forces (and thus potentially lower boiling points), it doesn't necessarily mean that the lower boiling point is exclusively due to the nonpolar nature of the liquid's molecules. Modified statement (b): The lower-boiling liquid could consist of nonpolar molecules or have weaker intermolecular forces.
03

Statement (c) Analysis and verification

Statement (c): The lower-boiling liquid has a lower molecular weight than the higher-boiling liquid. This statement is not always true. While it is generally correct that, all else being equal, a smaller molecular weight will result in a lower boiling point due to weaker intermolecular forces, there are exceptions. For example, hydrogen bonding between strongly polar molecules can result in higher boiling points even for molecules with lower molecular weights. Modified statement (c): The lower-boiling liquid may have a lower molecular weight or weaker intermolecular forces than the higher-boiling liquid.
04

Statement (d) Analysis and verification

Statement (d): The two liquids have identical vapor pressures at their normal boiling points. This statement is correct. At their normal boiling points, both liquids will have a vapor pressure of 760 mm Hg, which is the atmospheric pressure in this case since they are both boiling at atmospheric pressure.
05

Statement (e) Analysis and verification

Statement (e): At \( -84^{\circ}\) both liquids have vapor pressures of 760 \(\mathrm{mm} \mathrm{Hg}\). This statement is incorrect. The given value is only true for the liquid with a boiling point of \( -84^{\circ}\). At this temperature, the other liquid will have a vapor pressure lower than 760 mm Hg, as its boiling point is higher. Modified statement (e): At \( -84^{\circ}\), the liquid with a boiling point of \( -84^{\circ}\) has a vapor pressure of 760 \(\mathrm{mm} \mathrm{Hg}\), while the other liquid has a lower vapor pressure.

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Key Concepts

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

Intermolecular Forces
Understanding how intermolecular forces influence the boiling points of liquids is crucial for comprehending the physical properties of substances. Fundamentally, intermolecular forces are the forces of attraction and repulsion between molecules. They include dipole-dipole interactions, hydrogen bonding, and dispersion forces, also known as London forces.

These forces dictate how tightly molecules are held together in a liquid. A substance with stronger intermolecular forces will require more energy—in the form of heat—to break these attractions and transition from the liquid to the gaseous state, which leads to a higher boiling point. Conversely, weaker intermolecular forces mean that less energy is needed for the molecules to escape into the vapor phase, resulting in a lower boiling point.
Vapor Pressure
The concept of vapor pressure is a core piece in the puzzle of understanding a liquid's boiling point. Vapor pressure is the pressure exerted by the vapor in equilibrium with its liquid phase at a given temperature. The presence of stronger intermolecular forces can reduce a substance's vapor pressure because fewer molecules have the kinetic energy needed to escape into the vapor phase at a given temperature.

A liquid boils when its vapor pressure equals the external pressure, typically atmospheric pressure. Thus, substances with differing boiling points will have different vapor pressures at a given temperature. For example, at a liquid's boiling point, the vapor pressure will be 760 mm Hg, but at any other temperature, the vapor pressures will not be identical unless both substances are at their respective boiling points.
Molecular Weight
In the context of boiling points and intermolecular forces, molecular weight is another key factor. Generally, molecules with higher molecular weights have more electrons, leading to stronger dispersion forces due to greater polarizability. This typically results in higher boiling points. However, it is not the sole factor determining a substance's boiling point. The type of intermolecular forces (such as hydrogen bonding) can play a far more significant role.

Therefore, while a higher molecular weight can be associated with a higher boiling point through stronger intermolecular forces, it is not an absolute rule. Other types of intermolecular attractions can complicate this picture, as can the shape of the molecule and its ability to pack in the liquid phase.

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Most popular questions from this chapter

The fluorocarbon compound \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{F}_{3}\) has a normal boiling point of \(47.6^{\circ} \mathrm{C}\) . The specific heats of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{F}_{3}(l)\) and \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{F}_{3}(g)\) are 0.91 and \(0.67 \mathrm{J} / \mathrm{g}-\mathrm{K}\) , respectively. The heat of vaporization for the compound is 27.49 \(\mathrm{kJ} / \mathrm{mol}\) . Calculate the heat required to convert 35.0 \(\mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{F}_{3}\) from a liquid at \(10.00^{\circ} \mathrm{C}\) to a gas at \(105.00^{\circ} \mathrm{C}\) .

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) melts at \(-114^{\circ} \mathrm{C}\) and boils at \(78^{\circ} \mathrm{C}\) . The enthalpy of fusion of ethanol is \(5.02 \mathrm{kJ} / \mathrm{mol},\) and its enthalpy of vaporization is 38.56 \(\mathrm{kJ} / \mathrm{mol}\) . The specific heats of solid and liquid ethanol are 0.97 and \(2.3 \mathrm{J} / \mathrm{g}-\mathrm{K},\) respectively. (a) How much heat is required to convert 42.0 \(\mathrm{g}\) of ethanol at \(35^{\circ} \mathrm{C}\) to the vapor phase at \(78^{\circ} \mathrm{C} ?(\mathbf{b})\) How much heat is required to convert the same amount of ethanol at \(-155^{\circ} \mathrm{C}\) to the vapor phase at \(78^{\circ} \mathrm{C} ?\)

Describe how a cholesteric liquid crystal phase differs from a nematic phase.

The boiling points, surface tensions, and viscosities of water and several alcohols are as shown below:(a) From ethanol to propanol to \(n\) -butanol the boiling points, surface tensions, and viscosities all increase. What is the reason for this increase? (b) How do you explain the fact that propanol and ethylene glycol have similar molecular weights (60 versus 62 amu), yet the viscosity of ethylene glycol is more than 10 times larger than propanol? (c) How do you explain the fact that water has the highest surface tension but the lowest viscosity?

In terms of the arrangement and freedom of motion of the molecules, how are the nematic liquid crystalline phase and an ordinary liquid phase similar? How are they different?

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