Problem 93 - Chemistry: The Central Science

Question

Acetonitrile (CH \(_{3} \mathrm{CN}\) ) is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a \(1.80 \mathrm{M}\) LiBr solution in acetonitrile is \(0.826 \mathrm{~g} / \mathrm{cm}^{3}\). Calculate the concentration of the solution in (a) molality, (b) mole fraction of LiBr, (c) mass percentage of \(\mathrm{CH}_{3} \mathrm{CN}\).

Step-by-Step Solution

Answer

The molality of the LiBr solution in acetonitrile is approximately 2.69 mol/kg, the mole fraction of LiBr is approximately 0.10, and the mass percentage of CH3CN in the solution is approximately 81.07%.

Step 1: Calculate the mass of 1 liter solution

First, we will determine the mass of 1 liter (1000 cm^3) of the solution using the given density: Mass of the solution = Density × Volume Mass of the solution = \(0.826 \frac{g}{cm^3}\) × 1000 cm^3 = 826 g

Step 2: Calculate the moles of LiBr in 1 liter solution

Next, we will determine the moles of LiBr present in 1 liter of the solution using the given molarity: Moles of LiBr = Molarity × Volume Moles of LiBr = 1.80 mol/L × 1 L = 1.80 mol

Step 3: Calculate the mass of LiBr in the solution

Now, we will determine the mass of LiBr present in the solution by multiplying the moles and the molar mass of LiBr: Mass of LiBr = Moles of LiBr × Molar mass of LiBr Mass of LiBr = 1.80 mol × (6.939 g/mol + 79.904 g/mol) ≈ 1.80 mol × 86.84 g/mol ≈ 156.31 g

Step 4: Calculate the mass of CH3CN in the solution

We will now calculate the mass of CH3CN in the solution by subtracting the mass of LiBr from the total mass of the solution: Mass of CH3CN = Mass of the solution - Mass of LiBr Mass of CH3CN = 826 g - 156.31 g ≈ 669.69 g

Step 5: Calculate the molality of the solution

We will now calculate the molality (mol/kg) of the solution using the formula: Molality = \(\frac{\text{moles of solute}}{\text{mass of solvent (kg)}}\) Molality = \(\frac{1.80 \text{mol}}{0.66969 \text{kg}}\) ≈ 2.69 mol/kg

Step 6: Calculate the mole fraction of LiBr

To calculate the mole fraction of LiBr, we need to find the moles of CH3CN in the solution. We do this by dividing the mass of CH3CN by its molar mass: Moles of CH3CN = \(\frac{\text{mass of CH3CN}}{\text{molar mass of CH3CN}}\) Moles of CH3CN = \(\frac{669.69 \text{g}}{41.05 \text{g/mol}}\) ≈ 16.31 mol Now, we can determine the mole fraction of LiBr: Mole fraction of LiBr = \(\frac{\text{moles of LiBr}}{\text{total moles}}\) Mole fraction of LiBr = \(\frac{1.80 \text{mol}}{1.80 \text{mol} + 16.31 \text{mol}}\) ≈ 0.10

Step 7: Calculate the mass percentage of CH3CN

Finally, we can calculate the mass percentage of CH3CN in the solution: Mass percentage of CH3CN = \(\frac{\text{mass of CH3CN}}{\text{total mass}} \times 100\%\) Mass percentage of CH3CN = \(\frac{669.69 \text{g}}{826 \text{g}} \times 100\%\) ≈ 81.07% (a) The molality of the solution is approximately 2.69 mol/kg. (b) The mole fraction of LiBr in the solution is approximately 0.10. (c) The mass percentage of CH3CN in the solution is approximately 81.07%.

Key Concepts

Molality

Molality is a measure of the concentration of a solute in a solution, defined as the number of moles of solute per kilogram of solvent. It's symbolized as 'm' and its unit is mol/kg. Unlike molarity, which is sensitive to temperature changes due to volume expansion or contraction, molality remains constant because mass is unaffected by temperature.

When calculating molality, remember that it only considers the mass of the solvent, not the total mass of the solution. This feature makes it convenient for studies involving boiling point elevation or freezing point depression, as these colligative properties depend on the number of solute particles in a solvent, regardless of volume.

For example, if a solution contains 1.80 moles of a solute and the mass of the solvent (not the total solution) is 0.66969 kilograms, then the molality is calculated as \(\frac{1.80 \text{ mol}}{0.66969 \text{ kg}}\approx 2.69 \text{ mol/kg}\). Understanding this concept can significantly improve your solution concentration calculations.

Mole Fraction

The mole fraction is another way of expressing the concentration of a component in a mixture. It is the ratio of the number of moles of one component to the total number of moles of all components in the mixture. The mole fraction is unitless and is important in mixing rules for predicting the properties of the mixture.

To calculate the mole fraction, you need to know the moles of each component in your mixture. It's a straightforward computation: for component A, the mole fraction \(X_A\) is \(X_A = \frac{n_A}{n_A + n_B}\), where \(n_A\) is the moles of component A, and \(n_B\) is the moles of component B. For instance, if you have 1.80 moles of LiBr and 16.31 moles of CH3CN, the mole fraction of LiBr is \(\frac{1.80}{1.80 + 16.31} \approx 0.10\).

By understanding mole fraction, students can better grasp the concept of the solution being involved in a larger system where the proportions of each component are critical.

Mass Percentage

Mass percentage, often simply called percent concentration, is another common method to express the concentration of a solution. It tells you the mass of the solute compared to the total mass of the solution, expressed as a percentage. This is calculated by dividing the mass of the solute by the total mass of the solution, and then multiplying by 100%.

For example, when given a mixture of CH3CN and LiBr, if you've calculated the mass of CH3CN to be 669.69 g and the total mass of the solution as 826 g, you'd compute the mass percentage as \(\frac{669.69 \text{ g}}{826 \text{ g}} \times 100\% \approx 81.07\%\). This is especially useful in pharmacy to determine medication dosages and in chemistry to prep solutions. Students frequently encounter mass percentage when dealing with real-world applications such as the formulation of consumer products.

Highlighting these specifics can help students not just solve the problems, but also understand the real-life relevance of solution concentrations.