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(a) What is the molality of a solution formed by dissolving \(1.25\) mol of \(\mathrm{KCl}\) in \(16.0\) mol of water? (b) How many grams of sulfur \(\left(\mathrm{S}_{8}\right)\) must be dissolved in \(100.0 \mathrm{~g}\) naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) to make a \(0.12 \mathrm{~m}\) solution?

Short Answer

Expert verified
(a) To calculate the molality of the KCl solution, we first need to find the mass of water in kg; Mass of water = (16.0 mol) 脳 (18.015 g/mol) / 1000. Then, using the molality formula, m = (1.25 mol of KCl) / (mass of water in kg), we can find the molality of the solution. (b) To find the grams of S鈧 necessary for the 0.12 m solution, we first convert the mass of C鈧佲個H鈧 to kg: Mass of C鈧佲個H鈧 in kg = 100.0 g / 1000. Then, find the moles of S鈧 by rearranging the molality formula: moles of S鈧 = 0.12 m 脳 (mass of C鈧佲個H鈧 in kg). Finally, we convert moles of S鈧 to grams using the molar mass of S鈧 (256.52 g/mol): Mass of S鈧 = (moles of S鈧) 脳 (256.52 g/mol).

Step by step solution

01

(a) Determine the molality formula

Molality (m) is the number of moles of solute dissolved per kilogram of solvent. The formula for molality is: m = (moles of solute) / (mass of solvent in kg)
02

(a) Calculate the molality of the KCl solution

Given that 1.25 mol of KCl is dissolved in 16.0 mol of water, we can calculate the molality of the solution with the formula: m = (moles of KCl) / (mass of water in kg) Since we are given the moles of water (16.0 mol), we need to convert it to mass in kilograms. The molar mass of water (H鈧侽) is approximately 18.015 g/mol. Therefore, the mass of 16.0 mol of water is: Mass of water = (16.0 mol) 脳 (18.015 g/mol) To convert grams to kilograms, divide by 1000: Mass of water in kg = Mass of water / 1000 Now, we can calculate the molality of the solution: m = (1.25 mol of KCl) / (mass of water in kg)
03

(b) Understand how to convert molality to grams of solute

Given the molality of the b solution, 0.12 m, and that we want to find the grams of 饾憜鈧 necessary, we'll use the molality formula: m = (moles of S鈧) / (mass of C鈧佲個H鈧 in kg) We are given the mass of C鈧佲個H鈧 (100.0 g). First, we need to convert it to kilograms: Mass of C鈧佲個H鈧 in kg = Mass of C鈧佲個H鈧 / 1000 Now, we can rearrange the formula for molality to solve for moles of 饾憜鈧: moles of S鈧 = m 脳 (mass of C鈧佲個H鈧 in kg)
04

(b) Calculate the moles of S鈧 needed

Use the provided molality (0.12 m) and the calculated mass of C鈧佲個H鈧 in kg to find the moles of S鈧 necessary: moles of S鈧 = 0.12 m 脳 (mass of C鈧佲個H鈧 in kg)
05

(b) Convert moles of S鈧 to grams of S鈧

We now have the moles of 饾憜鈧 necessary for the solution. To find the mass in grams, we will use the molar mass of sulfur (S鈧). The molar mass of S鈧 is approximately 256.52 g/mol. Using this, we can convert moles of 饾憜鈧 to grams: Mass of S鈧 = (moles of S鈧) 脳 (256.52 g/mol)

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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 measure of how much solute is present in a given quantity of solvent or solution. In different chemical contexts, solution concentration can be expressed in various ways:
  • Molality (m): This is a measure of concentration expressed as moles of solute per kilogram of solvent. It is symbolized by the letter "m." Molality is particularly useful in scenarios where temperature changes, since it doesn't change with temperature because it's based on mass, not volume.
  • Molarity (M): This is another measure of concentration expressed as moles of solute per liter of solution. Unlike molality, molarity can change with temperature because solution volume can expand or contract depending on temperature.
  • Weight percent (wt%): This is the mass percentage of solute present in a mixture. It is the mass of the solute divided by the total mass of the solution, multiplied by 100.
For our exercise, we are focusing on determining the molality of a solution, which involves the moles of solute and mass of the solvent, as per the given chemical problems.
moles of solute
The concept of moles of solute is fundamental to calculating solution concentration. **Moles** represent a standard unit in chemistry that quantifies an amount of substance. One mole of any substance contains Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\) entities (atoms, molecules, etc.). This makes moles a convenient unit for counting atoms or molecules at the microscale.

For the first part of the exercise, dissolving \(1.25\) moles of potassium chloride (KCl) into water involves understanding that there are \(1.25\) moles of KCl molecules dissolved evenly throughout the solution. This makes it inherently relatable to find out how concentrated the solution is when paired with the amount of solvent (in this case, water) to receive the measure in terms of molality.

The moles of solute determine the number of chemical entities you are working with and directly influence solution properties like freezing point depression and boiling point elevation, commonly studied in colligative properties.
mass of solvent
When calculating the molality of a solution, understanding the mass of the solvent is critical. The **solvent** is the substance that dissolves the solute, and its mass plays a vital role in determining concentration in terms of molality.
  • Mass of solvent required: This is usually given in grams or kilograms. In our exercise, the conversion of solvent mass in grams to kilograms is necessary for using the molality formula.
  • Why kg?: Kilograms are used because molality is defined as moles of solute per kilogram of solvent. The unit of kilograms ensures consistency, since molality remains unaffected by temperature changes as it's based on mass and not volume.
  • Conversions: In the exercise provided, you'll need to convert the given moles of water (if not directly given as mass) to mass using the molar mass of water which is \(18.015 \text{ g/mol}\). Afterwards, further convert to kilograms by dividing by \(1000\).
Mass of the solvent helps provide meaningful context to the concentration of the solution and is indispensable in deriving the final solution's properties.

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Most popular questions from this chapter

Which of the following in each pair is likely to be more soluble in water: (a) cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) or glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) (Figure 13.12); (b) propionic acid \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right)\) or sodium propionate \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COONa}\right) ;\) (c) \(\mathrm{HCl}\) or ethyl chloride \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\right) ?\) Explain in each case.

Two beakers are placed in a sealed box at \(25^{\circ} \mathrm{C}\). One beaker contains \(30.0 \mathrm{~mL}\) of a \(0.050 \mathrm{M}\) aqueous solution of a nonvolatile nonelectrolyte. The other beaker contains \(30.0 \mathrm{~mL}\) of a \(0.035 \mathrm{M}\) aqueous solution of \(\mathrm{NaCl}\). The water vapor from the two solutions reaches equilibrium. (a) In which beaker does the solution level rise, and in which one does it fall? (b) What are the volumes in the two beakers when equilibrium is attained, assuming ideal behavior?

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

A \(40.0 \%\) by weight solution of \(\mathrm{KSCN}\) in water at \(20^{\circ} \mathrm{C}\) has a density of \(1.22 \mathrm{~g} / \mathrm{mL}\). (a) What is the mole fraction of \(\mathrm{KSCN}\) in the solution, and what are the molarity and molality? (b) Given the calculated mole fraction of salt in the solution, comment on the total number of water molecules available to hydrate each anion and cation. What ion pairing (if any) would you expect to find in the solution? Would you expect the colligative properties of such a solution to be those predicted by the formulas given in this chapter? Explain.

List four properties of a solution that depend on the total concentration but not the type of particle or particles present as solute. Write the mathematical expression that describes how each of these properties depends on concentration.

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