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Chapter 13: Problem 46

(a) What is the molality of a solution formed by dissolving 1.12 mol of KCl in 16.0 mol of water? (b) How many grams of sulfur \(\left(S_{8}\right)\) must be dissolved in \(100.0 \mathrm{~g}\) of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) to make a \(0.12 \mathrm{~m}\) solution?

Short Answer

Expert verified
(a) The molality of the KCl solution is 3.885 m. (b) 3.079 g of \(S_8\) are needed for the solution.

Step by step solution

01

Understanding Molality

The molality (m) of a solution is defined as the number of moles of solute per kilogram of solvent. It's important to note that molality uses kilograms of the solvent, not the solution.
02

Calculating Molality for KCl in Water

Given: 1.12 mol of KCl and 16.0 mol of water. First, convert moles of water to mass. Water has a molar mass of approximately 18.02 g/mol. Thus, 16.0 mol of water: \(16.0 \text{ mol} \times 18.02 \, \text{g/mol} = 288.32 \, \text{g} \). Convert grams to kilograms: 288.32 g = 0.28832 kg. Molality = moles of solute / kilograms of solvent = \( \frac{1.12}{0.28832} = 3.885 \, m \).
03

Understanding Molality for the Second Part

For the sulfur in naphthalene, you need to find the amount of solute required to achieve a specific molality. Given the desired molality, we find how much solute is needed using grams of solvent.
04

Using Molality Formula to Find Grams for S in Naphthalene

Given: Desired molality (m) = 0.12 m, mass of naphthalene = 100.0 g = 0.1 kg, formula of sulfur \(S_8\). The formula connects molality with moles and kilograms of solvent: \(m = \frac{\text{moles of } S_8}{0.1} \). Therefore, moles of \(S_8 = 0.12 \times 0.1 = 0.012 \). Next, find the mass of \(S_8\): Molar mass of \(S_8\) = \(8 \times 32.07 = 256.56 \) g/mol. Thus, mass of \(S_8 = 0.012 \times 256.56 = 3.07872 \) grams.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solute
In chemistry, a solute is the substance that is dissolved in a solvent to form a solution. The solute is present in a smaller proportion compared to the solvent. The essence of a solution revolves around how the solute interacts with the solvent.

When calculating the molality of a solution, the solute's amount, given in moles, plays a crucial role. For example, if you have a solution of potassium chloride (KCl) in water, KCl is the solute.
  • The formula for molality involves the molar amount of the solute.
  • In our exercise, 1.12 moles of KCl were the moles of solute used to measure molality.
  • The importance lies in ensuring the solute amount is accurately calculated in moles for precise solution concentration.
Understanding and calculating the moles of your solute is foundational to solving many chemistry problems.
Solvent
The solvent is the component in a solution that dissolves the solute. It is generally present in a larger amount compared to the solute. In a solution, the solvent's primary role is to disperse the solute particles.

In the context of molality, solvent mass directly impacts the calculation:
  • Molality is determined by dividing the moles of solute by the mass of solvent in kilograms.
  • In the first part of the exercise, water acts as the solvent, with 16.0 moles translating to 288.32 grams (0.28832 kilograms).
  • Keeping track of the mass in kilograms is crucial because molality is expressed per kilogram of solvent, not per the entire solution.
Remember, understanding the characteristics of your solvent helps in achieving accurate and meaningful molarity or molality values.
Moles
Moles are a fundamental unit of measurement in chemistry, representing a quantity of chemical entities (atoms, molecules, etc.). The concept of moles simplifies dealing with chemical reactions and solutions.

When discussing solutions and molarity or molality, knowing how to convert between moles and other units is vital:
  • In the exercise, moles were used to measure both the solute (KCl and sulfur, \(S_8\)) and the solvent (water).
  • The conversion from moles to grams (and vice versa) is important, especially when dealing with molar mass. For instance, 16.0 moles of water is converted to grams using its molar mass, 18.02 g/mol.
  • Similarly, the exercise required calculating moles of sulfur based on desired molality, which was then converted to grams using sulfur's molar mass.
Having a solid grasp of calculating moles and using them in formulae ensures successful outcomes in chemistry problem-solving.

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Most popular questions from this chapter

Fluorocarbons (compounds that contain both carbon and fluorine) were, until recently, used as refrigerants. The compounds listed in the following table are all gases at \(25^{\circ} \mathrm{C},\) and their solubilities in water at \(25^{\circ} \mathrm{C}\) and 101.3 kPa fluorocarbon pressure are given as mass percentages. (a) For each fluorocarbon, calculate the molality of a saturated solution. (b) Which molecular property best predicts the solubility of these gases in water: molar mass, dipole moment, or ability to hydrogen-bond to water? (c) Infants born with severe respiratory problems are sometimes given liquid ventilation: They breathe a liquid that can dissolve more oxygen than air can hold. One of these liquids is a fluorinated compound, \(\mathrm{CF}_{3}\left(\mathrm{CF}_{2}\right)_{7} \mathrm{Br}\). The solubility of oxygen in this liquid is \(66 \mathrm{~mL} \mathrm{O}_{2}\) per \(100 \mathrm{~mL}\) liquid. In contrast, air is \(21 \%\) oxygen by volume. Calculate the moles of \(\mathrm{O}_{2}\) present in an infant's lungs (volume: \(15 \mathrm{~mL}\) ) if the infant takes a full breath of air compared to taking a full "breath" of a saturated solution of \(\mathrm{O}_{2}\) in the fluorinated liquid. Assume a pressure of \(101.3 \mathrm{kPa}\) in the lungs. $$ \begin{array}{lc} \hline \text { Fluorocarbon } & \text { Solubility (mass \%) } \\ \hline \mathrm{CF}_{4} & 0.0015 \\ \mathrm{CClF}_{3} & 0.009 \\ \mathrm{CCl}_{2} \mathrm{~F}_{2} & 0.028 \\ \mathrm{CHClF}_{2} & 0.30 \\ \hline \end{array} $$

Commercial concentrated aqueous ammonia is \(28 \% \mathrm{NH}_{3}\) by mass and has a density of \(0.90 \mathrm{~g} / \mathrm{mL}\). What is the molarity of this solution?

The solubility of alum, \(\mathrm{KAl}\left(\mathrm{SO}_{4}\right)_{2} \cdot 12 \mathrm{H}_{2} \mathrm{O},\) in water at is \(44 \mathrm{~g}\) per \(100 \mathrm{~g}\) of water at \(50^{\circ} \mathrm{C}\). A solution of alum in water at \(80^{\circ} \mathrm{C}\) is formed by dissolving \(130 \mathrm{~g}\) in \(100 \mathrm{~g}\) of water. When this solution is slowly cooled to \(50^{\circ} \mathrm{C}\), no precipitate forms. (a) Is the solution that has cooled down to \(50^{\circ} \mathrm{C}\) unsaturated, saturated, or supersaturated? (b) You take a metal spatula and scratch the side of the glass vessel that contains this cooled solution, and crystals start to appear. What has just happened? (c) At equilibrium, what mass of crystals do you expect to form?

The following table presents the solubilities of several gases in water at \(25^{\circ} \mathrm{C}\) under a total pressure of gas and water vapor of \(101.3 \mathrm{kPa}\). (a) What volume of \(\mathrm{CH}_{4}(g)\) under standard conditions of temperature and pressure is contained in \(4.0 \mathrm{~L}\) of a saturated solution at \(25^{\circ} \mathrm{C} ?\) (b) The solubilities (in water) of the hydrocarbons are as follows: methane \(<\) ethane \(<\) ethylene. Is this because ethylene is the most polar molecule? (c) What intermolecular interactions can these hydrocarbons have with water? (d) Draw the Lewis dot structures for the three hydrocarbons. Which of these hydrocarbons possess \(\pi\) bonds? Based on their solubilities, would you say \(\pi\) bonds are more or less polarizable than \(\sigma\) bonds? (e) Explain why \(\mathrm{NO}\) is more soluble in water than either \(\mathrm{N}_{2}\) or \(\mathrm{O}_{2}\). (f) \(\mathrm{H}_{2} \mathrm{~S}\) is more water-soluble than almost all the other gases in table. What intermolecular forces is \(\mathrm{H}_{2} \mathrm{~S}\) likely to have with water? \((\mathbf{g}) \mathrm{SO}_{2}\) is by far the most water-soluble gas in table. What intermolecular forces is \(\mathrm{SO}_{2}\) likely to have with water? $$ \begin{array}{lc} \hline \text { Gas } & \text { Solubility (mM) } \\ \hline \mathrm{CH}_{4} \text { (methane) } & 1.3 \\ \mathrm{C}_{2} \mathrm{H}_{6} \text { (ethane) } & 1.8 \\ \mathrm{C}_{2} \mathrm{H}_{4} \text { (ethylene) } & 4.7 \\ \mathrm{~N}_{2} & 0.6 \\ \mathrm{O}_{2} & 1.2 \\ \mathrm{NO} & 1.9 \\ \mathrm{H}_{2} \mathrm{~S} & 99 \\ \mathrm{SO}_{2} & 1476 \\ \hline \end{array} $$

The Henry's law constant for hydrogen gas \(\left(\mathrm{H}_{2}\right)\) in water at \(25^{\circ} \mathrm{C}\) is \(7.7 \times 10^{-6} \mathrm{M} / \mathrm{kPa}\) and the constant for argon \((\mathrm{Ar})\) at \(25^{\circ} \mathrm{C}\) is \(1.4 \times 10^{-5} \mathrm{M} / \mathrm{kPa}\). If the two gases are each present at 253 kPa pressure, calculate the solubility of each gas.
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