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Chapter 11: Problem 97

The solubility of benzoic acid, is 0.34 \(\mathrm{g} / 100 \mathrm{mL}\) in water at \(25^{\circ} \mathrm{C}\) and 10.0 \(\mathrm{g} / 100 \mathrm{mL}\) in benzene $\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\( at \)25^{\circ} \mathrm{C} .$ Rationalize this solubility behavior. For a \(1.0-\mathrm{m}\) solution of benzoic acid in benzene, would the measured freezing point depression be equal to, greater than, or less than $5.12^{\circ} \mathrm{C} ?\left(K_{\mathrm{f}}=5.12^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol} \text { for benzene.) }\right.$

Short Answer

Expert verified
The solubility of benzoic acid is higher in benzene than in water because benzoic acid is an organic compound and more soluble in nonpolar solvents like benzene. For a 1.0 m solution of benzoic acid in benzene, the measured freezing point depression would be equal to 5.12 °C, as calculated using the formula ΔT = Kf × m × i, with i=1 for benzoic acid in benzene.

Step by step solution

01

Analyze the nature of the substances and their interactions

Benzoic acid is an organic compound, while water is a polar compound due to the presence of the hydroxyl group (-OH). Benzene is also an organic compound, which makes it nonpolar. According to the principle "like dissolves like", polar compounds are more soluble in polar solvents and nonpolar compounds in nonpolar solvents. This explains why benzoic acid has higher solubility in benzene.
02

Calculate the expected freezing point depression

To determine whether the measured freezing point depression would be equal to, greater than, or less than \(5.12^{\circ} \mathrm{C}\), we will use the freezing point depression formula: ΔT = Kf × m × i where ΔT represents the freezing point depression, Kf is the cryoscopic constant for the solvent, m is the molality of the solution, and i is the dissociation factor. Since benzoic acid is a weak acid and does not dissociate into ions in a nonpolar solvent like benzene, i = 1.
03

Plug in the values and calculate the freezing point depression

Given a 1.0 m solution of benzoic acid in benzene, and using the Kf value provided for benzene, we can plug in the values into the formula: ΔT = (5.12 °C · kg/mol) × (1.0 mol/kg) × (1) ΔT = 5.12 °C
04

Compare the calculated freezing point depression to the given value of 5.12 °C

Since the calculated freezing point depression is exactly equal to the provided value of 5.12 °C, we conclude that the measured freezing point depression for a 1.0 m solution of benzoic acid in benzene would be equal to 5.12 °C.

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Most popular questions from this chapter

Which ion in each of the following pairs would you expect to be more strongly hydrated? Why? a. \(\mathrm{Na}^{+}\) or \(\mathrm{Mg}^{2+}\) b. \(\mathrm{Mg}^{2+}\) or \(\mathrm{Be}^{2+}\) c. \(\mathrm{Fe}^{2+}\) or \(\mathrm{Fe}^{3+}\) d. \(\mathrm{F}^{-}\) or \(\mathrm{Br}^{-}\) e. \(\mathrm{Cl}^{-}\) or \(\mathrm{ClO}_{4}^{-}\) f. \(\mathrm{ClO}_{4}^{-}\) or \(\mathrm{SO}_{4}^{2-}\)

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