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 \(\mathrm{CH}_{3} \mathrm{CN}\).

Short answer

The concentration of the \(1.80\,\text{M}\) LiBr solution in acetonitrile is (a) 2.69 mol/kg in molality, (b) 0.0995 in mole fraction of LiBr, and (c) 81.08% mass percentage of \(\mathrm{CH}_{3} \mathrm{CN}\).

Step-by-step solution

Find moles and volume of the solution using given molarity

The molarity of the LiBr solution is given as \(1.80\,\text{M}\). We can use the definition of molarity to find the moles of solute and volume of the solution. Let's assume we have 1.00 L of the solution. Moles of solute (LiBr) = molarity × volume of solution Moles of LiBr = \(1.80\,\text{M} \times 1.00\,\text{L} = 1.80\,\text{mol}\)

Find the mass of the solution using density

Given, the density of the solution is \(0.826\,\text{g}/\text{cm}^3\). We can use this to find the mass of the 1.00 L solution: 1.00 L = 1000 cm³ Mass = Density × volume Mass of solution = \(0.826\,\text{g}/\text{cm}^3 \times 1000\,\text{cm}^3 = 826.0\,\text{g}\)

Calculate the mass of LiBr and acetonitrile

We know the moles of LiBr are 1.80 mol. To find the mass, we need to use the molar mass of LiBr, which is: Molar mass of LiBr = 6.94 (Li) + 79.90 (Br) = 86.84 g/mol Mass of LiBr = moles × molar mass Mass of LiBr = 1.80 mol × 86.84 g/mol = 156.31 g Next, we can calculate the mass of the solvent (acetonitrile, \(\mathrm{CH}_{3} \mathrm{CN}\)): Mass of acetonitrile = mass of solution - mass of LiBr Mass of acetonitrile = 826.0 g - 156.31 g = 669.69 g

Calculate the molality

Now, we can find the molality using the definition: molality = moles of solute / mass of solvent (kg) molality = 1.80 mol / (669.69 g × 0.001 kg/g) = 2.69 mol / kg The molality of the LiBr solution is 2.69 mol / kg.

Calculate the mole fraction

First, we need to find the moles of acetonitrile. To do this, we will use the molar mass of \(\mathrm{CH}_{3} \mathrm{CN}\) (41.05 g/mol): Moles of acetonitrile = mass / molar mass Moles of acetonitrile = 669.69 g / 41.05 g/mol = 16.31 mol Now, we can calculate the mole fraction of LiBr: Mole fraction of LiBr = moles of LiBr / (moles of LiBr + moles of acetonitrile) Mole fraction of LiBr = 1.80 mol / (1.80 mol + 16.31 mol) The mole fraction of LiBr is 0.0995.

Calculate the mass percentage of acetonitrile

Finally, we can calculate the mass percentage of \(\mathrm{CH}_{3} \mathrm{CN}\): Mass percentage of \(\mathrm{CH}_{3} \mathrm{CN}\) = (mass of acetonitrile / mass of solution) × 100% Mass percentage of \(\mathrm{CH}_{3} \mathrm{CN}\) = (669.69 g / 826.0 g) × 100% The mass percentage of \(\mathrm{CH}_{3} \mathrm{CN}\) is 81.08%. In summary, the concentration of the 1.80 M LiBr solution in acetonitrile is (a) 2.69 mol / kg in molality, (b) 0.0995 in mole fraction of LiBr, and (c) 81.08% mass percentage of \(\mathrm{CH}_{3} \mathrm{CN}\).

Key concepts

Molality

Molality is a way to express the concentration of a solution, focusing on the moles of solute per kilogram of solvent. This measurement is independent of temperature or pressure since it relies on mass rather than volume.

To calculate molality, use the formula:
\[ \text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
For example, given a 1.80 M LiBr solution with acetonitrile, where the solvent's mass is 669.69 g, we find:

• Convert the mass of the solvent to kilograms: 669.69 g = 0.66969 kg.
• Use the formula: \( \frac{1.80\, \text{mol}}{0.66969\, \text{kg}} \approx 2.69\, \text{mol/kg} \).

This means the solution's concentration is 2.69 mol/kg.

Understanding molality is crucial for assessing solutions' behavior in varying conditions.

Mole Fraction

Mole fraction quantifies the ratio of moles of a component to the total moles in the solution. It provides a unitless value that ranges from 0 to 1, indicating the proportion of each substance within the mixture.

For the calculation of the mole fraction of LiBr in the solution:

• First, calculate the moles of acetonitrile, given its mass of 669.69 g and molar mass 41.05 g/mol: \( \frac{669.69\, \text{g}}{41.05\, \text{g/mol}} = 16.31\, \text{mol} \).
• Next, use the formula for mole fraction of LiBr:
  \[ \text{Mole fraction of LiBr} = \frac{\text{moles of LiBr}}{\text{moles of LiBr} + \text{moles of acetonitrile}} \]
  \[ \frac{1.80\, \text{mol}}{1.80\, \text{mol} + 16.31\, \text{mol}} \approx 0.0995 \].

This value tells us that LiBr makes up approximately 9.95% of the moles in this solution.

Mole fraction is helpful for understanding the composition of mixtures in chemistry.

Mass Percentage

Mass percentage represents the ratio of the mass of a constituent to the total mass of the solution, expressed as a percentage. It's a practical way to convey how much of a solute or solvent is present.

To find the mass percentage of acetonitrile in our solution:

• Use the mass of acetonitrile (669.69 g) and the total mass of the solution (826.0 g).
• Apply the formula:
  \[ \text{Mass percentage of } \mathrm{CH}_3\mathrm{CN} = \left(\frac{\text{mass of acetonitrile}}{\text{total mass of solution}}\right) \times 100\% \]
• Plug in the values: \[ \left(\frac{669.69\, \text{g}}{826.0\, \text{g}}\right) \times 100\% \approx 81.08\% \].

This means acetonitrile comprises 81.08% of the solution by mass.

Mass percentage is often used because it gives a clear picture of concentration in practical terms.