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Chapter 9: Problem 40

Calculate the concentrations of each of the following solutions: (a) the molality of \(13.63 \mathrm{~g}\) of sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), dissolved in \(612 \mathrm{~mL}\) of water; (b) the molality of CsCl in a \(10.00 \%\) by mass aqueous solution; (c) the molality of acetone in an aqueous solution with a mole fraction for acetone of \(0.197 .\)

Short Answer

Expert verified
The molalities are: (a) 0.0651 m, (b) 0.66 m, (c) approximately 0.44 m.

Step by step solution

01

Calculate the molar mass of sucrose

To find the molality, we need to find the molar mass of sucrose, which is \(C_{12}H_{22}O_{11}\). Add up the atomic masses of all the atoms in the molecule: \(12 \times \text{Carbon}) + (22 \times \text{Hydrogen}) + (11 \times \text{Oxygen}) = 12 \times 12.01 + 22 \times 1.01 + 11 \times 16.00 = 342.34 \text{g/mol}\).
02

Convert grams of sucrose to moles

Using the molar mass from Step 1, convert the mass of sucrose to moles: \(\frac{13.63 \text{g}}{342.34 \text{g/mol}} = 0.03982 \text{moles}\).
03

Convert volume of water to kilograms

To find molality, we need the mass of solvent in kilograms. Since 1 mL of water has a mass of 1 g, \(612 \text{mL}\) of water is equivalent to \(612 \text{g} = 0.612 \text{kg}\).
04

Calculate the molality of sucrose solution

Molality \(m\) is defined as moles of solute per kilogram of solvent. From Steps 2 and 3, we have: \(m = \frac{0.03982 \text{moles}}{0.612 \text{kg}} = 0.0651 \text{mol/kg}\).
05

Calculate molality of CsCl from a 10% by mass solution

For a \(10.00\text{%}\) by mass solution, we have \(100 \text{g}\) of solution containing \(10 \text{g}\) of CsCl and \(90 \text{g}\) of water (\(0.090 \text{kg}\)). The molar mass of CsCl is \(Cs + Cl = 132.91 + 35.45 = 168.36 \text{g/mol}\). The moles of CsCl is \(\frac{10 \text{g}}{168.36 \text{g/mol}} = 0.0594 \text{moles}\). The molality is \(m = \frac{0.0594 \text{moles}}{0.090 \text{kg}} = 0.66 \text{mol/kg}\).
06

Calculate molality of acetone from mole fraction

Mole fraction (\(\chi\)) is the ratio of moles of one component to the total moles in the solution. If acetone's mole fraction is \(0.197\), then water's mole fraction is \(1 - 0.197 = 0.803\). Assuming 1 kg of solution, since the density of such mixed solutions is close to water, the moles of water (\(\text{molar mass} = 18.02 \text{ g/mol}\)) will be \(\text{moles of water} = 0.803 \times 1000 \/ 18.02\). The moles of acetone will be \(0.197 \times 1000 \/ 18.02\). Then, molality (\(m\)) of acetone will be the ratio of moles of acetone to the mass of water in kilograms: \(m = \frac{0.197 \times 1000 \/ 58.08}{0.803 \times 1000 \/ 18.02} = 0.44 \text{mol/kg}\).

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Key Concepts

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

Molar Mass
Understanding molar mass is essential for calculating solution concentrations. It represents the mass of one mole (or 6.022 x 1023 particles) of any substance and is expressed in grams per mole (g/mol). To calculate the molar mass, sum the atomic masses of all atoms in the molecule, as provided by the periodic table.

For instance, in our exercise, the molar mass of sucrose (\(C_{12}H_{22}O_{11}\)) is calculated by adding the atomic mass of carbon (12.01 g/mol) 12 times, hydrogen (1.01 g/mol) 22 times, and oxygen (16.00 g/mol) 11 times. This yields a molar mass of 342.34 g/mol for sucrose, which is critical for the subsequent molality calculation.
Moles Conversion
The concept of moles conversion is fundamental to chemistry. It allows us to convert between the mass of a substance and the number of moles, facilitating various calculations involving reactions and solution concentrations.

To convert grams to moles, divide the mass of the substance by its molar mass. In the given exercise, we've converted 13.63 g of sucrose to moles. With the molar mass of sucrose at 342.34 g/mol, the calculation is \(\frac{13.63 \text{g}}{342.34 \text{g/mol}} = 0.03982 \text{moles}\). This is a pivotal step when we need to find the molality of a solution.
Solution Concentration
Solution concentration can be quantified in several ways, including molarity and molality. Molality, in particular, is the measure of the number of moles of solute per kilogram of solvent, which makes it temperature-independent and valuable for various scientific analyses.

To calculate molality, divide the number of moles of solute by the mass of solvent in kilograms. As illustrated in the sucrose example, the molality is \(m = \frac{0.03982 \text{moles}}{0.612 \text{kg}} = 0.0651 \text{mol/kg}\), which represents a precise way to express the concentration of the sucrose in water.
Mole Fraction
The mole fraction is another important concept for understanding solution compositions. It is the ratio of the moles of a particular component to the total moles of all substances in the solution. The mole fraction is unitless and provides insight into the composition of mixtures.

For example, with an acetone mole fraction of 0.197, we can calculate the remaining fraction for water and use these proportions to calculate the molality of acetone in the solution. Understanding mole fractions is crucial for accurately describing the makeup of solutions in various chemical contexts.

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Most popular questions from this chapter

Complete the following statements about the effect of intermolecular forces on the physical properties of a substance. (a) The higher the boiling point of a liquid, the (stronger, weaker) are its intermolecular forces. (b) Substances with strong intermolecular forces have (high, low) vapor pressures. (c) Substances with strong intermolecular forces typically have (high, low) surface tensions. (d) The higher the vapor pressure of a liquid, the (stronger, weaker) are its intermolecular forces. (c) Because nitrogen, \(\mathrm{N}_{2}\), has (strong, weak) intermolecular forces, it has a (high, low) critical temperature. (f) Substances with high vapor pressures have correspondingly (high, low) boiling points. (g) Because water has a high boiling point, it must have (strong, weak) intermolecular forces and a correspondingly (high, low) enthalpy of vaporization.

Explain how the vapor pressure of a liquid is affected by each of the following changes in conditions: (a) an increase in temperature; (b) an increase in surface area of the liquid; (c) an increase in volume abowe the liquid; (d) the addition of air to the volume above the liquid.

Distinguish between a foam and a sol. Give at least one example of each.

The carbon dioxide gas dissolved in a sample of water in a partly filled, sealed container has reached equilibrium with irs partial pressure in the air above the solution. Explain what happens to the solubility of the \(\mathrm{CO}_{2}\) if (a) the partial pressure of the \(\mathrm{CO}_{2}\) gas is doubled by the addition of more \(\mathrm{CO}_{2}\); (b) the total pressure of the gas above the liquid is doubled by the addirion of nitrogen.

The volume of blood in the body of a certain deep-sea diver is about \(6.00 \mathrm{~L}\). Blood cells make up about \(55 \%\) of the blood volume, and the remaining \(45 \%\) is the aqueous solution called plasma. What is the maximum volume of nitrogen measured at \(1.00 \mathrm{~atm}\) and \(37^{\circ} \mathrm{C}\) that could dissolve in the diver's blood plasma at a depth of \(93 \mathrm{~m}\), where the pressure is \(10.0\) atm? (This is the volume that could come out of solution suddenly, causing the painful and dangerous condition called the bends, if the diver were to ascend too quickly.) Assume that Henry's constant for nitrogen at \(37^{\circ} \mathrm{C}\) (body temperature) is \(5.8 \times 10^{-4} \mathrm{~mol} \cdot \mathrm{L}^{-1}\)-atm \({ }^{-1}\).
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