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Chapter 10: Problem 112

For each of the following pairs, predict which substance is more soluble in water. a. \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) or \(\mathrm{NH}_{3}\) b. \(\mathrm{CH}_{3} \mathrm{CN}\) or \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) d. \(\mathrm{CH}_{3} \mathrm{OH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) e. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH}\) or \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{OH}\) f. \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) or \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\)

Short Answer

Expert verified
a. \(\mathrm{NH}_{3}\) is more soluble. b. \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) is more soluble. c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) is more soluble. d. \(\mathrm{CH}_{3} \mathrm{OH}\) is more soluble. e. \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{OH}\) is more soluble. f. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) is more soluble.

Step by step solution

01

a. Comparing \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) and \(\mathrm{NH}_{3}\)

Both of these substances are polar and will engage in hydrogen bonding with water. While both will be soluble, \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) has an additional methyl group which is nonpolar and may cause a slight decrease in solubility. Therefore, \(\mathrm{NH}_{3}\) is more soluble in water.
02

b. Comparing \(\mathrm{CH}_{3} \mathrm{CN}\) and \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\)

The polarity of \(\mathrm{CH}_{3} \mathrm{CN}\) is due to the cyano group, while the polarity of \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) is due to hydrogen bonding with the oxygen. In this case, \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) is more soluble in water due to stronger hydrogen bonding.
03

c. Comparing \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\)

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) is polar due to the hydroxyl group, which can engage in hydrogen bonding with water. On the other hand, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) is nonpolar and has only weak van der Waals forces. Therefore, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) is more soluble in water.
04

d. Comparing \(\mathrm{CH}_{3} \mathrm{OH}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)

Both of these substances are polar due to the hydroxyl group. However, \(\mathrm{CH}_{3} \mathrm{OH}\) has a smaller nonpolar region than \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\), making it slightly more soluble in water.
05

e. Comparing \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH}\) and \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{OH}\)

Both of these substances have a polar hydroxyl group capable of hydrogen bonding. However, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH}\) has a larger nonpolar region than \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{OH}\), which decreases its solubility in water. Therefore, \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{OH}\) is more soluble in water.
06

f. Comparing \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) and \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\)

Both of these substances are polar. \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) engages in hydrogen bonding with water, while \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) can form hydrogen bonds and also dissociate into ions. In this case, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) is more soluble in water due to its ability to dissociate into ions, which further increases the solubility.

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Most popular questions from this chapter

A solution is prepared by mixing \(50.0 \mathrm{mL}\) toluene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3},\right.\) \(d=0.867 \mathrm{g} / \mathrm{cm}^{3}\) with \(125 \mathrm{mL}\) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}, d=0.874 \mathrm{g} / \mathrm{cm}^{3}\right)\) Assuming that the volumes add on mixing, calculate the mass percent, mole fraction, molality, and molarity of the toluene.

A solution is prepared by dissolving 52.3 g cesium chloride in \(60.0 \mathrm{g}\) water. The volume of the solution is \(63.3 \mathrm{mL}\). Calculate the mass percent, molarity, molality, and mole fraction of the CsCl solution.

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You have a solution of two volatile liquids, \(A\) and \(B\) (assume ideal behavior). Pure liquid A has a vapor pressure of 350.0 torr and pure liquid B has a vapor pressure of 100.0 torr at the temperature of the solution. The vapor at equilibrium above the solution has double the mole fraction of substance A that the solution does. What is the mole fraction of liquid A in the solution?

Plants that thrive in salt water must have internal solutions (inside the plant cells) that are isotonic with (have the same osmotic pressure as) the surrounding solution. A leaf of a saltwater plant is able to thrive in an aqueous salt solution (at \(25^{\circ} \mathrm{C}\) ) that has a freezing point equal to \(-0.621^{\circ} \mathrm{C}\). You would like to use this information to calculate the osmotic pressure of the solution in the cell. a. In order to use the freezing-point depression to calculate osmotic pressure, what assumption must you make (in addition to ideal behavior of the solutions, which we will assume)? b. Under what conditions is the assumption (in part a) reasonable? c. Solve for the osmotic pressure (at \(25^{\circ} \mathrm{C}\) ) of the solution in the plant cell. d. The plant leaf is placed in an aqueous salt solution (at \(\left.25^{\circ} \mathrm{C}\right)\) that has a boiling point of \(102.0^{\circ} \mathrm{C}\). What will happen to the plant cells in the leaf?
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