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

Short Answer

Expert verified
a) To find the molality of the KCl solution, first, convert the moles of water to mass (in kg) using the molar mass of water (18.02 g/mol): Mass of water (kg) = (16.0 mol 脳 18.02 g/mol) 梅 1000 = 0.288 kg Then, calculate the molality of the KCl solution using the formula: Molality (m) = 1.12 mol 梅 0.288 kg = 3.89 m b) To find the mass of S鈧 needed, first, convert the mass of naphthalene to kg: Mass of naphthalene (kg) = 100.0 g 梅 1000 = 0.1 kg Then, find the moles of S鈧 using the given molality: Moles of S鈧 = 0.12 m 脳 0.1 kg = 0.012 mol Finally, convert the moles of S鈧 to mass using the molar mass of S鈧 (256.52 g/mol): Mass of S鈧 = 0.012 mol 脳 256.52 g/mol = 3.08 g

Step by step solution

01

Identify the given values

In this problem, we are given: - Moles of KCl = 1.12 mol - Moles of water = 16.0 mol
02

Convert moles of water to mass of water in kg

We know that the molar mass of water (H鈧侽) is 18.02 g/mol. Therefore, to find the mass of water, we can use the following equation: Mass of water = Moles of water 脳 Molar mass of water Now, we will convert this mass of water to kg: Mass of water (kg) = Mass of water (g) 梅 1000
03

Calculate the molality of the KCl solution

Now we can use the molality formula: Molality (m) = moles of KCl 梅 mass of water (kg) Plug in the values and find the molality of the KCl solution. #b) Find the mass of S鈧 needed to make a 0.12 m solution in 100.0 g of naphthalene#
04

Identify the given values

In this problem, we are given: - Molality (m) = 0.12 m - Mass of naphthalene (C鈧佲個H鈧) = 100.0 g
05

Convert mass of naphthalene to mass of naphthalene in kg and find moles of solute

To find the mass of naphthalene in kg, divide the given mass by 1000: Mass of naphthalene (kg) = Mass of naphthalene (g) 梅 1000 Next, use the molality formula to find the moles of S鈧 (solute): Molality (m) = moles of S鈧 梅 mass of naphthalene (kg) Rearrange the formula to find the moles of S鈧: Moles of S鈧 = Molality (m) 脳 mass of naphthalene (kg) Plug in the values to find the moles of S鈧.
06

Convert moles of S鈧 to mass of S鈧

We know that the molar mass of S鈧 (sulfur) is 256.52 g/mol. Therefore, to find the mass of S鈧, we can use the following equation: Mass of S鈧 = Moles of S鈧 脳 Molar mass of S鈧 Calculate the mass of S鈧 required to make a 0.12 m solution in 100.0 g of naphthalene.

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Key Concepts

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

Solution Concentration
Solution concentration refers to the amount of a solute that is dissolved in a given quantity of solvent. Understanding solution concentration is essential in chemistry, as it helps determine how concentrated or dilute a solution is. In this context, molality (\( m \)) is a useful measure of concentration, especially when temperature changes are involved since it is independent of temperature. Molality is defined as the number of moles of solute per kilogram of solvent.

To calculate the molality, first, the amount of solute in moles is determined, then the mass of the solvent is measured in kilograms. The formula is simple: \[ m = \frac{{\text{{moles of solute}}}}{{\text{{mass of solvent (kg)}}}} \] For instance, if you dissolve 1.12 moles of potassium chloride (\( \text{{KCl}} \) in 16.0 moles of water (\( \text{{H}}_2\text{{O}} \) the molality of the solution can be calculated using the steps given in the textbook solution.
Molar Mass
Molar mass is the mass of one mole of a substance and it is expressed in grams per mole (g/mol). It is a fundamental concept for converting between the mass of a substance and the number of moles. Each element has a unique molar mass, which is equivalent to its atomic weight listed on the periodic table. For compounds, the molar mass is the sum of the molar masses of all the atoms in the molecule.

For example, the molar mass of water (H鈧侽) is calculated by adding twice the molar mass of hydrogen (about 1.01 g/mol) and once the molar mass of oxygen (about 16.00 g/mol), giving us approximately 18.02 g/mol. Similarly, to find the mass of sulfur octamer (\( S_8 \) which is needed to create a solution, knowing that its molar mass is 256.52 g/mol allows the direct conversion from moles to grams and vice versa.
Moles to Mass Conversion
The process of converting moles to mass and vice versa is critical in preparing solutions of precise concentrations. This fundamental calculation requires the molar mass of the solute. To convert the number of moles of a substance to mass, you multiply the number of moles by the molar mass of the substance. The formula is: \[ \text{{Mass (g)}} = \text{{Moles}} \times \text{{Molar mass (g/mol)}} \] Conversely, to find the number of moles when you have the mass, you divide the mass by the molar mass: \[ \text{{Moles}} = \frac{{\text{{Mass (g)}}}}{{\text{{Molar mass (g/mol)}}}} \]

Using sulfur octamer (\( S_8 \) as an example from the original exercise, once its molar mass is known (256.52 g/mol), it is straightforward to convert the calculated moles of \( S_8 \) needed for the solution into mass by simply multiplying the moles by the molar mass. This is how, after calculating the molality of the solution, we find how many grams of \( S_8 \) are required to obtain the desired solution concentration.

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Most popular questions from this chapter

Indicate whether each statement is true or false: (a) NaCl dissolves in water but not in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) because benzene is denser than water. (b) NaCl dissolves in water but not in benzene because water has a large dipole moment and benzene has zero dipole moment. (c) NaCl dissolves in water but not in benzene because the water-ion interactions are stronger than benzene-ion interactions.

Seawater contains 34 g of salts for every liter of solution. Assuming that the solute consists entirely of \(\mathrm{NaCl}\) (in fact, over 90\(\%\) of the salt is indeed NaCl), calculate the osmotic pressure of seawater at \(20^{\circ} \mathrm{C}\) .

Describe how you would prepare each of the following aqueous solutions, starting with solid KBr: (a) 0.75 L of \(1.5 \times 10^{-2} M \mathrm{KBr},\) (b) 125 \(\mathrm{g}\) of \(0.180 \mathrm{m} \mathrm{KBr},(\mathbf{c}) 1.85 \mathrm{L}\) of a solution that is 12.0\(\% \mathrm{KBr}\) by mass (the density of the solution is 1.10 \(\mathrm{g} / \mathrm{mL}\) , ( \(\mathrm{d}\) ) a 0.150 \(\mathrm{M}\) solution of \(\mathrm{KBr}\) that contains just enough KBr to precipitate 16.0 \(\mathrm{g}\) of AgBr from a solution containing 0.480 mol of \(\mathrm{AgNO}_{3} .\)

A solution is made containing 14.6 \(\mathrm{g}\) of \(\mathrm{CH}_{3} \mathrm{OH}\) in 184 \(\mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) Calculate (a) the mole fraction of \(\mathrm{CH}_{3} \mathrm{OH},\) (b) the mass percent of \(\mathrm{CH}_{3} \mathrm{OH},(\mathbf{c})\) the molality of \(\mathrm{CH}_{3} \mathrm{OH}\) .

You make a solution of a nonvolatile solute with a liquid solvent. Indicate if each of the following statements is true or false. (a) The freezing point of the solution is unchanged by addition of the solvent. (b) The solid that forms as the solution freezes is nearly pure solute. (c) The freezing point of the solution is independent of the concentration of the solute. ( \(\mathbf{d}\) ) The boiling point of the solution increases in proportion to the concentration of the solute. (e) At any temperature, the vapor pressure of the solvent over the solution is lower than what it would be for the pure solvent.

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