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Chapter 13: Problem 5

Two of the substances listed here are highly soluble in water, two are only slightly soluble in water, and two are insoluble in water. Indicate the situation you expect for each one. (a) iodoform, \(\mathrm{CHI}_{3}\) (b) benzoic acid, (c) formic acid, (d) 1-butanol, (e) chlorobenzene, (f) propylene glycol, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}\)

Short Answer

Expert verified
The highly soluble substances are: formic acid, \( \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH} \). The slightly soluble substances are: iodoform, 1-butanol. Benzoic acid and chlorobenzene are insoluble.

Step by step solution

01

Analyzing Iodoform

Iodoform \(\mathrm{CHI}_{3}\) is a halogen compound. The iodine atoms make the molecule quite large and nonpolar, therefore, though it could have some aqueous solubility, its interaction with water will be limited. Hence, it is slightly soluble in water.
02

Analyzing Benzoic Acid

Benzoic acid is a carboxylic acid with a large non-polar region (benzene ring). However, it also has an -COOH group which can perform hydrogen bonding with water, hence it is slightly soluble in water despite being predominantly non-polar.
03

Analyzing Formic Acid

Formic acid is a simple carboxylic acid with two very polar functional groups: the carboxylic acid group and the hydroxyl group. These can easily form hydrogen bonds with water, enhancing its solubility. Thus, formic acid is highly soluble in water.
04

Analyzing 1-butanol

1-butanol has a polar -OH (alcohol) group that can form hydrogen bonds with water. However, the butyl group \(C_4H_9\) at one end of the molecule is quite large and non-polar. Hence, 1-butanol is slightly soluble in water.
05

Analyzing Chlorobenzene

Chlorobenzene is a large, predominantly non-polar molecule with a single polar bond due to the presence of Chlorine. Its ability to form hydrogen bonds with water is, however, quite limited. Thus, it is insoluble in water.
06

Analyzing Propylene Glycol

Propylene glycol \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}\) has two polar -OH groups that can form hydrogen bonds with water, making it highly soluble in water despite its size.

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Most popular questions from this chapter

Styrene, used in the manufacture of polystyrene plastics, is made by the extraction of hydrogen atoms from ethylbenzene. The product obtained contains about \(38 \%\) styrene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}=\mathrm{CH}_{2}\right)\) and \(62 \%\) ethylbenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}_{3}\right),\) by mass. The mixture is separated by fractional distillation at \(90^{\circ} \mathrm{C} .\) Determine the composition of the vapor in equilibrium with this \(38 \%-62 \%\) mixture at \(90^{\circ} \mathrm{C}\). The vapor pressure of ethylbenzene is \(182 \mathrm{mmHg}\) and that of styrene is \(134 \mathrm{mmHg}\).

The concentration of \(\mathrm{N}_{2}\) in the ocean at \(25^{\circ} \mathrm{C}\) is \(445 \mu \mathrm{M} .\) The Henry's law constant for \(\mathrm{N}_{2}\) is \(0.61 \times 10^{-3} \mathrm{mol} \mathrm{L}^{-1} \mathrm{atm}^{-1} .\) Calculate the mass of \(\mathrm{N}_{2}\) in a liter of ocean water. Calculate the partial pressure of \(\mathrm{N}_{2}\) in the atmosphere.

Natural gas consists of about \(90 \%\) methane, \(\mathrm{CH}_{4}\) Assume that the solubility of natural gas at \(20^{\circ} \mathrm{C}\) and 1 atm gas pressure is about the same as that of \(\mathrm{CH}_{4}\) \(0.02 \mathrm{g} / \mathrm{kg}\) water. If a sample of natural gas under a pressure of 20 atm is kept in contact with \(1.00 \times 10^{3} \mathrm{kg}\) of water, what mass of natural gas will dissolve?

Predict the approximate freezing points of \(0.10 \mathrm{m}\) solutions of the following solutes dissolved in water: (a) \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(\text { urea }) ;(\mathrm{b}) \mathrm{NH}_{4} \mathrm{NO}_{3} ;(\mathrm{c}) \mathrm{HCl} ;(\mathrm{d}) \mathrm{CaCl}_{2}\) (e) \(\mathrm{MgSO}_{4} ;\) (f) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (ethanol); \((\mathrm{g}) \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (acetic acid).

An aqueous solution is \(6.00 \%\) methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) by mass, with \(d=0.988 \mathrm{g} / \mathrm{mL} .\) What is the molarity of \(\mathrm{CH}_{3} \mathrm{OH}\) in this solution?
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