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Chapter 13: Problem 33

Which of the following in each pair is likely to be more soluble in hexane, \(\mathrm{C}_{6} \mathrm{H}_{14} :(\mathbf{a}) \mathrm{CCl}_{4}\) or \(\mathrm{CaCl}_{2},(\mathbf{b})\) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) or glycerol, \(\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH},(\mathbf{c})\) octanoic \(\mathrm{acid}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH},\) or acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH} ?\) Explain your answer in each case.

Short Answer

Expert verified
In hexane, the more soluble compounds in each pair are: (a) CCl4, (b) Benzene, and (c) Octanoic acid. This is because hexane is a nonpolar solvent, and the selected compounds have nonpolar characteristics or longer nonpolar chains, adhering to the "like dissolves like" principle.

Step by step solution

01

Analyzing Polarity of CCl4 and CaCl2

In pair (a), we have Carbon Tetrachloride (CCl4) and Calcium Chloride (CaCl2). CCl4 is a nonpolar molecule, as it has a symmetric distribution of the electron cloud. On the other hand, CaCl2 is an ionic compound, which is highly polar.
02

Comparing Solubilities of CCl4 and CaCl2 in Hexane

Since hexane is a nonpolar solvent and CCl4 is a nonpolar molecule, CCl4 is more likely to be soluble in hexane than the polar CaCl2.
03

Analyzing Polarity of Benzene and Glycerol

In pair (b), we have benzene (C6H6) and glycerol (CH2(OH)CH(OH)CH2OH). Benzene is a nonpolar molecule, with a symmetrical structure and a uniform distribution of the electron cloud. Glycerol, however, has three hydroxyl groups (-OH), which makes it a polar molecule.
04

Comparing Solubilities of Benzene and Glycerol in Hexane

Since hexane is a nonpolar solvent, benzene, which is also nonpolar, is more likely to be soluble in hexane compared to the polar glycerol.
05

Analyzing Polarity of Octanoic acid and Acetic acid

In pair (c), we have octanoic acid (CH3(CH2)6COOH) and acetic acid (CH3COOH). Both of these molecules are carboxylic acids, and they have polar regions due to the presence of the carboxyl group (-COOH). However, octanoic acid also has a long nonpolar chain of hydrocarbons.
06

Comparing Solubilities of Octanoic acid and Acetic acid in Hexane

Due to the longer nonpolar chain in octanoic acid, it is more likely to be soluble in the nonpolar solvent hexane compared to the smaller acetic acid molecule, which has a higher proportion of polar regions. In conclusion, the compounds that are more likely to be soluble in hexane are: (a) CCl4 (b) Benzene (c) Octanoic acid

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Most popular questions from this chapter

Describe how you would prepare each of the following aqueous solutions: (a) 1.50 \(\mathrm{L}\) of 0.110 \(\mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) solution, starting with solid \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} ;\) (b) 225 \(\mathrm{g}\) of a solution that is 0.65 \(\mathrm{m}\) in \(\mathrm{Na}_{2} \mathrm{CO}_{3},\) starting with the solid solute; ( c ) 1.20 L of a solution that is 15.0\(\% \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) by mass (the density of the solution is 1.16 \(\mathrm{g} / \mathrm{mL}\) , starting with solid solute; (\boldsymbol{d} ) ~ a ~ 0.50 \(\mathrm{M}\) solution of HCl that would just neutralize 5.5 \(\mathrm{g}\) of \(\mathrm{Ba}(\mathrm{OH})_{2}\) starting with 6.0 \(\mathrm{M} \mathrm{HCl}\) .

You make a solution of a nonvolatile solute with a liquid solvent. Indicate whether each of the following statements is true or false. (a) The freezing point of the solution is higher than that of the pure solvent. (b) The freezing point of the solution is lower than that of the pure solvent. (c) The boiling point of the solution is higher than that of the pure solvent. (d) The boiling point of the solution is lower than that of the pure solvent.

(a) Calculate the vapor pressure of water above a solution prepared by adding 22.5 g of lactose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) to 200.0 \(\mathrm{g}\) of water at 338 \(\mathrm{K}\) . (Vapor-pressure data for water are given in Appendix B.) (b) Calculate the mass of propylene glycol \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{2}\right)\) that must be added to 0.340 \(\mathrm{kg}\) of water to reduce the vapor pressure by 2.88 torr at \(40^{\circ} \mathrm{C}\) .

If you compare the solubilities of the noble gases in water, you find that solubility increases from smallest atomic weight to largest, Ar \(<\mathrm{Kr}<\mathrm{Xe} .\) Which of the following statements is the best explanation? [ Section 13.3\(]\) (a) The heavier the gas, the more it sinks to the bottom of the water and leaves room for more gas molecules at the top of the water. (b) The heavier the gas, the more dispersion forces it has, and therefore the more attractive interactions it has with water molecules. (c) The heavier the gas, the more likely it is to hydrogen-bond with water. (d) The heavier the gas, the more likely it is to make a saturated solution in water.

Aerosols are important components of the atmosphere. Does the presence of aerosols in the atmosphere increase or decrease the amount of sunlight that arrives at the Earth's surface, compared to an "aerosol-free" atmosphere? Explain your reasoning.
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