Problem 40 - Chemical Principles

Question

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 \(0.197 .\)

Step-by-Step Solution

Answer

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

Step 1: 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: \(12 \times \text{Carbon}) + (22 \times \text{Hydrogen}) + (11 \times \text{Oxygen}) = 12 \times 12.01 + 22 \times 1.01 + 11 \times 16.00 = 342.34 \text{g/mol}\).

Step 2: Convert grams of sucrose to moles

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

Step 3: 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 \(612 \text{g} = 0.612 \text{kg}\).

Step 4: 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: \(m = \frac{0.03982 \text{moles}}{0.612 \text{kg}} = 0.0651 \text{mol/kg}\).

Step 5: 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 \(m = \frac{0.0594 \text{moles}}{0.090 \text{kg}} = 0.66 \text{mol/kg}\).

Step 6: 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: \(m = \frac{0.197 \times 1000 \/ 58.08}{0.803 \times 1000 \/ 18.02} = 0.44 \text{mol/kg}\).

Key Concepts

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.