Problem 98 - Chemistry The Central Science

Question

Acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) 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 short answer to the given problem is as follows: (a) The molality of the LiBr solution is 2.688 mol/kg. (b) The mole fraction of LiBr in the solution is 0.09935. (c) The mass percentage of CH₃CN in the solution is 81.07%.

Step 1: Calculate the mass of 1 L of solution

To calculate the mass of 1 L solution, we use the density formula: Density = \( \frac{mass}{volume} \) Rearranging, mass = Density × volume Given, Density = 0.826 g/cm³. We need to convert this density to g/L by multiplying with \(1000^3\). So, the density = 0.826 g/cm³ × 1000³ = 826 g/L. Now, we will find the mass of 1L solution: mass = Density × volume = 826 g/L × 1 L = 826 g

Step 2: Calculate the mass of LiBr in 1 L of solution

To find the mass of the solute, LiBr, in 1 L of solution, we will use the given concentration of the LiBr solution: Concentration (M) = \( \frac{moles\,of\, solute}{volume\, of\, solution\, in\, L} \) Rearranging, moles of solute = Concentration × volume Given, Concentration = 1.80 M, volume = 1 L Now, we will find the moles of LiBr: moles of LiBr = 1.80 M × 1 L = 1.80 moles Next, we will find the molar mass of LiBr: Molar mass of LiBr (Li = 6.939 g/mol, Br = 79.904 g/mol) = 86.843 g/mol Now, we will find the mass of LiBr: mass of LiBr = moles of LiBr × molar mass of LiBr = 1.80 moles × 86.843 g/mol = 156.32 g

Step 3: Calculate the mass of CH3CN in 1 L of solution

To find the mass of the solvent, CH3CN, we will subtract the mass of solute, LiBr from the total mass of the solution: mass of CH3CN = mass of solution - mass of LiBr = 826 g - 156.32 g = 669.68 g

Step 4: Calculate the molality of the solution

To find the molality of the solution, we use the formula: Molality = \( \frac{moles\, of\, solute}{mass\, of\, solvent\, in\, kg} \) First, convert the mass of solvent, CH3CN to kg: mass of CH3CN in kg = 669.68 g × 0.001 = 0.66968 kg Molality = \( \frac{1.80\, moles\, of\, LiBr}{0.66968\, kg\, CH3CN} \) = 2.688 mol/kg (a) The molality of the solution is 2.688 mol/kg.

Step 5: Calculate the mole fraction of LiBr

To find the mole fraction of LiBr, we use the formula: Mole fraction of LiBr = \( \frac{moles\, of\, LiBr}{total\, moles\, of\, solute\, and\, solvent} \) First, find the moles of CH3CN in the solution: Molar mass of CH₃CN = (12.01 + (3 × 1.01) + 14.01 + 12.01 g/mol)= 41.05 g/mol Moles of CH₃CN = \( \frac{mass\, of\, CH3CN}{molar\, mass\, of\, CH3CN} \) = \( \frac{669.68\, g}{41.05\, g/mol} \) = 16.316 moles Total moles of solute and solvent = moles of LiBr + moles of CH₃CN = 1.80 moles + 16.316 moles = 18.116 moles Mole fraction (χ) of LiBr = \( \frac{1.80}{18.116} \) = 0.09935 (b) The mole fraction of LiBr is 0.09935.

Step 6: Calculate the mass percentage of CH3CN

To find the mass percentage of the solvent, CH3CN, we will use the formula: Mass percentage of CH₃CN = \( \frac{mass\, of\, CH3CN}{total\, mass\, of\, solution} \) × 100 Mass percentage of CH₃CN = \( \frac{669.68\, g}{826\, g} \) × 100 = 81.07 % (c) The mass percentage of CH3CN in the solution is 81.07 %.

Key Concepts

Molality

Molality is a way to express the concentration of a solution. It measures the number of moles of solute (the substance dissolved) per kilogram of solvent (the substance doing the dissolving). Unlike molarity, molality doesn't change with temperature because it depends on mass, not volume.

To calculate molality, use the formula:

Molality (m) = \( \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)

This approach is particularly useful in scenarios like boiling-point elevation and freezing-point depression. In our exercise, we found the molality of a LiBr solution in acetonitrile to be 2.688 mol/kg. This involves converting the mass of the solvent into kilograms and dividing the moles of the solute by this mass.

Mole Fraction

The mole fraction is another way to express concentration, focusing on the ratio of moles of a specific component to the total moles of all components in the mixture. It’s a unitless measure and is always less than one.

For calculations:

Mole fraction of a component \( = \frac{\text{moles of component}}{\text{total moles of all components}} \)

This concept shows the proportion of a certain substance in the mixture. It is useful because it remains constant regardless of temperature or pressure changes. In our example, we calculated that the mole fraction of LiBr in the solution was 0.09935, indicating a small proportion of LiBr relative to the acetonitrile and LiBr combined.

Mass Percentage

Mass percentage is a simple, yet effective way to express concentration that describes the mass of a specific component relative to the total mass of the mixture, expressed as a percentage.

For the formula:

Mass percentage \( = \frac{\text{mass of component}}{\text{total mass of solution}} \times 100 \)

Mass percentage helps in understanding the composition of mixtures in a straightforward way. In our scenario, the mass percentage of acetonitrile in the solution was found to be 81.07%, which shows that acetonitrile makes up the majority of the solution by mass.

Density Calculations

Density is the mass per unit volume of a substance and is a crucial factor in converting between mass and volume. It is especially valuable when working with concentrations that rely on the total volume of a solution.

Utilize the formula:

Density = \( \frac{\text{mass}}{\text{volume}} \)

This makes it simple to calculate the mass of a specific volume of solution, which can then be used in further concentration calculations as seen in the previous sections. In our exercise, we converted the density of the LiBr solution to find out that 1 liter of the solution weighs 826 g, aiding in determining other concentration measures.