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?

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: \[ \text{distance ratio} = \sqrt{\frac{(1/1.08)}{(1/1.266)}} \]

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.

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.