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Benzoic acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH},\) melts at \(122^{\circ} \mathrm{C}\). The density in the liquid state at \(130^{\circ} \mathrm{C}\) is \(1.08 \mathrm{~g} / \mathrm{cm}^{3} .\) The density of solid benzoic acid at \(15^{\circ} \mathrm{C}\) is \(1.266 \mathrm{~g} / \mathrm{cm}^{3}\). (a) In which of these two states is the average distance between molecules greater? (b) If you converted a cubic centimeter of liquid benzoic acid into a solid, would the solid take up more, or less, volume than the original cubic centimeter of liquid?

Short Answer

Expert verified
(a) The average distance between molecules is greater in the liquid state of benzoic acid. (b) The solid benzoic acid takes up less volume than the original cubic centimeter of liquid.

Step by step solution

01

Calculate the average distance ratio between liquid and solid states

The ratio of the average distance between molecules in the two states can be calculated as a square root of the ratio of the inverse densities: \[ \text{distance ratio} = \sqrt{\frac{(1/\rho_\text{liquid})}{(1/\rho_\text{solid})}} \] where \(\rho_\text{liquid}\) and \(\rho_\text{solid}\) are the densities of benzoic acid in the liquid and solid states, respectively. We are given the following density values: - Liquid density: \(\rho_\text{liquid} = 1.08 \, \mathrm{g/cm^3}\) - Solid density: \(\rho_\text{solid} = 1.266 \, \mathrm{g/cm^3}\) Now we can calculate the distance ratio: \[ \text{distance ratio} = \sqrt{\frac{(1/1.08)}{(1/1.266)}} \]
02

Determine which state has a greater average distance between molecules

Calculate the distance ratio: \[ \text{distance ratio} = \sqrt{\frac{(1/1.08)}{(1/1.266)}} \approx 1.08 \] Since the distance ratio is greater than 1, the average distance between molecules is greater in the liquid state of benzoic acid.
03

Determine the volume change when converting between states

Consider a 1 cm³ sample of liquid benzoic acid. To find the mass of this sample, use the density formula with liquid density: \[ m_\text{liquid} = \rho_\text{liquid} V_\text{liquid} = 1.08 \, \mathrm{g/cm^3} \times 1 \, \mathrm{cm^3} = 1.08 \, \mathrm{g} \] When converting to solid, the mass remains constant, which means \(m_\text{solid} = m_\text{liquid} = 1.08 \, \mathrm{g}\). Now use the density formula once again, this time for the solid state, to find the volume of the solid benzoic acid: \[ V_\text{solid} = \frac{m_\text{solid}}{\rho_\text{solid}} = \frac{1.08 \, \mathrm{g}}{1.266 \, \mathrm{g/cm^3}} \approx 0.85 \, \mathrm{cm^3} \] Since \(V_\text{solid} < V_\text{liquid}\), the solid benzoic acid takes up less volume than the original cubic centimeter of liquid. In conclusion, (a) The average distance between molecules is greater in the liquid state of benzoic acid. (b) The solid benzoic acid takes up less volume than the original cubic centimeter of liquid.

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Key Concepts

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

Molecular Distance
In chemistry, understanding the molecular distance is crucial as it relates to how molecules are packed in different states of matter. Molecular distance essentially refers to the average space between molecules in a material. In the context of benzoic acid, when it transitions from solid to liquid, the molecular distance changes. This change can be calculated by examining the densities of benzoic acid in its different states.

When benzoic acid is in its liquid form, the density is lower, suggesting that molecules are further apart. This is supported by our calculation where we found the distance ratio between liquid and solid benzoic acid to be about 1.08.
  • Liquid density: 1.08 g/cm³
  • Solid density: 1.266 g/cm³
The greater distance ratio indicates that molecules in liquid benzoic acid are spaced out more than in its solid form, showcasing increased molecular distance.
Volume Change
Volume change is an important concept when substances undergo phase transitions, such as melting or freezing. The overall space a substance occupies can alter dramatically depending on its state. This concept comes into play when considering 1 cm³ of liquid benzoic acid and its phase transition to a solid.

The exercise asks if the volume would increase or decrease upon solidification. By keeping the mass constant and observing that 1 cm³ of liquid turns into only about 0.85 cm³ of solid, it is evident that a phase shift to solid results in a reduction of volume.
  • Mass (liquid and solid remains same): 1.08 g
  • Liquid volume initially: 1 cm³
  • Solid volume: approximately 0.85 cm³
This decrease in volume shows how densely the molecules pack in the solid state, highlighting the volume change during such a transition.
Density Calculation
Calculating density is a fundamental skill for understanding how mass relates to volume. The concept of density (\( \rho \)) is defined as mass per unit volume, expressed typically in grams per cubic centimeter (g/cm³). In the case of benzoic acid, we are comparing its density in liquid and solid forms.

To calculate density or use it for further computations, you need to know either two of three variables: density, mass, and volume. The formula is straightforward:
\[ \rho = \frac{m}{V} \]
where \( m \) is mass and \( V \) is volume.
  • Density of liquid benzoic acid: 1.08 g/cm³
  • Density of solid benzoic acid: 1.266 g/cm³
In this problem, we used the liquid density to understand the mass of a 1 cm³ sample and predicted what volume it would occupy once it becomes solid. The increased density in the solid state confirmed that it occupies less volume, validating the calculations and helping understand its density behavior better.

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Most popular questions from this chapter

As a metal such as lead melts, what happens to (a) the average kinetic energy of the atoms and (b) the average distance between the atoms?

(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?

Appendix \(\mathrm{B}\) lists the vapor pressure of water at various external pressures. (a) Plot the data in Appendix B,vapor pressure 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 \(101.3 \mathrm{kPa}, 100^{\circ} \mathrm{C} .(\mathbf{c}) \mathrm{A}\) city at an altitude of \(1525 \mathrm{~m}\) above sea level has a barometric pressure of \(84.3 \mathrm{kPa}\). To what temperature would you have to heat water to boil it in this city? (d) A city at an altitude of \(150 \mathrm{~m}\) below sea level would have a barometric pressure of \(103.14 \mathrm{kPa}\). To what temperature would you have to heat water to boil it in this city?

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

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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