/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 Suppose you have two colorless m... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose you have two colorless molecular liquids \(A\) and \(B\) whose boiling points are \(78^{\circ} \mathrm{C}\) and \(112^{\circ} \mathrm{C}\) respectively and both are 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) Both A and B are liquids with identical vapor pressure at room temperature of \(25^{\circ} \mathrm{C} .(\mathbf{b})\) Liquid \(\mathrm{A}\) must consist of nonpolar molecules with lower molecular weight than B. \((\mathbf{c})\) Both liquids \(A\) and \(B\) have higher total intermolecular forces than water. (d) Liquid \(A\) is more volatile than liquid B because it has a lower boiling point. (e) At \(112^{\circ} \mathrm{C}\) both liquids have a vapor pressure of 1 atm.

Short Answer

Expert verified
(d) is correct; all other statements need modification as discussed.

Step by step solution

01

Understand Boiling Point Concept

Boiling point of a liquid is the temperature at which its vapor pressure equals the external pressure. Since atmospheric pressure is 1 atm, the liquids A and B will have vapor pressures equal to 1 atm at their respective boiling points (78°C for A and 112°C for B).
02

Analyze Statement (a)

Statement (a) claims both A and B have identical vapor pressures at 25°C. This is incorrect; vapor pressure decreases with an increase in boiling point. Modify the statement: 'Liquid A has a higher vapor pressure than liquid B at 25°C.'
03

Analyze Statement (b)

Statement (b) says liquid A must consist of nonpolar molecules with a lower molecular weight than B. This is partly incorrect; lower molecular weight can influence the boiling point, but nonpolarity isn't a necessity. Modify the statement: 'Liquid A likely has a lower molecular weight than B but may or may not be nonpolar.'
04

Analyze Statement (c)

Statement (c) asserts both A and B have higher intermolecular forces than water. This is incorrect; since both have lower boiling points than water (100°C), their intermolecular forces are likely weaker. Modify the statement: 'Both liquids A and B may have weaker intermolecular forces compared to water.'
05

Analyze Statement (d)

Statement (d) indicates liquid A is more volatile due to its lower boiling point. This is correct because volatility is inversely related to boiling point; the lower the boiling point, the higher the volatility.
06

Analyze Statement (e)

Statement (e) suggests at 112°C both liquids have a vapor pressure of 1 atm. This is incorrect for liquid A because it's already boiling at 78°C. Modify the statement: 'At 112°C, liquid B has a vapor pressure of 1 atm, while liquid A's vapor pressure would exceed 1 atm.'

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Key Concepts

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

Vapor Pressure
Vapor pressure is a crucial concept when studying the boiling points of liquids. It refers to the pressure exerted by a vapor in equilibrium with its liquid phase. When the vapor pressure equals the external atmospheric pressure, the liquid boils. The notion of vapor pressure explains why different substances have different boiling points.

For instance, liquid A with a boiling point of 78°C reaches an atmospheric pressure of 1 atm at that temperature, indicating a higher vapor pressure at lower temperatures compared to liquid B, which boils at 112°C. As a result, at any temperature below their boiling points, liquid A typically has a higher vapor pressure than liquid B, making it crucial in determining their relative volatilities. This discrepancy arises due to differences in molecular interactions within the liquids.
Intermolecular Forces
Intermolecular forces determine much about a liquid's properties, including its boiling point. These forces are the attractions between molecules. The stronger the intermolecular forces, the more energy is needed to separate the molecules and, thus, the higher the boiling point.

Both liquids A and B have lower boiling points than water (100°C), suggesting they have weaker intermolecular forces compared to water. Intermolecular forces include various types such as hydrogen bonding, dipole-dipole interactions, and London dispersion forces. For example, if one of the liquids exhibits hydrogen bonding, it would generally have a higher boiling point compared to another liquid with only London dispersion forces, assuming all else is equal.
  • Hydrogen bonding is the strongest among these and often leads to higher boiling points.
  • London dispersion forces are the weakest, influenced strongly by molecular weight and shape.
  • Dipole-dipole interactions fall intermediate in strength.
These forces play a vital role in determining the thermal behavior of substances.
Volatility
Volatility describes a liquid's tendency to evaporate. It's closely related to vapor pressure—a higher vapor pressure at a given temperature means a substance is more volatile. Therefore, volatility is inversely related to boiling point: the lower the boiling point, the higher the volatility.

Liquid A, with a boiling point of 78°C, is more volatile than liquid B, which boils at 112°C. As liquid A requires less heat to reach its vapor pressure equivalent to atmospheric pressure, it tends to evaporate more readily than liquid B. This explains why volatile liquids are often associated with high vapor pressures, as they can turn into vapor under less intense thermal conditions.
Molecular Weight
Molecular weight plays a significant role in determining physical properties such as boiling points and vapor pressures. It refers to the mass of a molecule, and generally, lighter molecules with lower molecular weights tend to have lower boiling points.

In the case of liquids A and B, liquid A having a lower boiling point implies it might have a lower molecular weight compared to B. This is because lighter molecules can generate greater kinetic energy at lower temperatures, increasing their vapor pressure, hence boiling earlier. However, the impact of molecular weight is not universal; structural aspects and intermolecular forces also significantly influence these properties. Thus, while liquid A may likely be lighter, its exact nature depends on the interplay of all these factors together.

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Most popular questions from this chapter

The table shown here lists the molar heats of vaporization for several organic compounds. Use specific examples from this list to illustrate how the heat of vaporization varies with (a) molar mass, (b) molecular shape, \((\mathbf{c})\) molecular polarity, (d) hydrogen-bonding interactions. Explain these comparisons in terms of the nature of the intermolecular forces at work. (You may find it helpful to draw out the structural formula for each compound.) \begin{tabular}{lc} \hline Compound & Heat of Vaporization (kJ/mol) \\ \hline \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) & 19.0 \\ \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) & 27.6 \\ \(\mathrm{CH}_{3} \mathrm{CHBrCH}_{3}\) & 31.8 \\ \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) & 32.0 \\ \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}\) & 33.6 \\ \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) & 47.3 \\ \hline \end{tabular}

For a given substance, the liquid crystalline phase tends to be more viscous than the liquid phase. Why?

Due to the environmental concern of fluorocarbons as refrigerants, a refrigerant based on a mixture of hydrocarbons was used as a replacement. It is a patented blend of ethane, propane, butane, and isobutane. Isobutane has a normal boiling point of \(-12^{\circ} \mathrm{C}\). The molar specific heat of liquid phase and gas phase isobutane are \(129.7 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) and \(95.2 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) respectively. The heat of vaporization for this compound is \(21.3 \mathrm{~kJ} / \mathrm{mol}\). Calculate the heat required to convert \(25.0 \mathrm{~g}\) of isobutane from a liquid at \(-50^{\circ} \mathrm{C}\) to a gas at \(40^{\circ} \mathrm{C}\).

(a) When you exercise vigorously, you sweat. How does this help your body cool? (b) A flask of water is connected to a vacuum pump. A few moments after the pump is turned on, the water begins to boil. After a few minutes, the water begins to freeze. Explain why these processes occur.

Suppose the vapor pressure of a substance is measured at two different temperatures. (a) By using the Clausius-Clapeyron equation (Equation 11.1\()\) derive the following relationship between the vapor pressures, \(P_{1}\) and \(P_{2}\), and the absolute temperatures at which they were measured, \(T_{1}\) and \(T_{2}\) : $$ \ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\text {vap }}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) $$ (b) Gasoline is a mixture of hydrocarbons, a component of which is octane \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\right)\). Octane has a vapor pressure of \(1.86 \mathrm{kPa}\) at \(25^{\circ} \mathrm{C}\) and a vapor pressure of \(19.3 \mathrm{kPa}\) at \(75^{\circ} \mathrm{C}\). Use these data and the equation in part (a) to calculate the heat of vaporization of octane. \((\mathbf{c})\) By using the equation in part (a) and the data given in part (b), calculate the normal boiling point of octane. Compare your answer to the one you obtained from Exercise 11.81 . (d) Calculate the vapor pressure of octane at \(-30^{\circ} \mathrm{C}\).

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