Problem 49
Question
The density of acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) is \(0.786 \mathrm{~g} / \mathrm{mL}\) and the density of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). A solution is made by dissolving \(25.0 \mathrm{~mL}\) of \(\mathrm{CH}_{3} \mathrm{OH}\) in \(100 \mathrm{~mL}\) of \(\mathrm{CH}_{3} \mathrm{CN}\). (a) What is the mole fraction of methanol in the solution? (b) What is the molality of the solution? (c) Assuming that the volumes are additive, what is the molarity of \(\mathrm{CH}_{3} \mathrm{OH}\) in the solution?
Step-by-Step Solution
Verified Answer
Mole fraction: 0.244. Molality: 7.85 mol/kg. Molarity: 4.94 mol/L.
1Step 1: Calculate mass of methanol
Use the density and volume of methanol to find its mass. Given: Density of methanol \(= 0.791 \text{ g/mL}\), Volume \(= 25.0 \text{ mL}\). Thus, \( \text{mass} = \text{density} \times \text{volume} = 0.791 \times 25.0 = 19.775 \text{ g}\).
2Step 2: Calculate mass of acetonitrile
Use the density and volume of acetonitrile to find its mass. Given: Density of acetonitrile \(= 0.786 \text{ g/mL}\), Volume \(= 100 \text{ mL}\). Thus, \( \text{mass} = \text{density} \times \text{volume} = 0.786 \times 100 = 78.6 \text{ g}\).
3Step 3: Convert masses to moles
Find the molar mass of each compound and convert the mass to moles. Methanol: \( \text{CH}_3\text{OH} = 12.01 + 3(1.01) + 16.00 + 1.01 = 32.04 \text{ g/mol}\). Moles of methanol \( = \frac{19.775}{32.04} \approx 0.617 \text{ mol}\). Acetonitrile: \( \text{CH}_3\text{CN} = 12.01 + 3(1.01) + 12.01 + 14.01 = 41.05 \text{ g/mol}\). Moles of acetonitrile \( = \frac{78.6}{41.05} \approx 1.914 \text{ mol}\).
4Step 4: Calculate mole fraction of methanol
Mole fraction of methanol \(\chi_{\text{CH}_3\text{OH}}\) is calculated using the formula: \(\chi_{\text{CH}_3\text{OH}} = \frac{\text{moles of methanol}}{\text{moles of methanol} + \text{moles of acetonitrile}} = \frac{0.617}{0.617 + 1.914} \approx 0.244\).
5Step 5: Calculate molality of methanol
Molality is calculated using moles of solute and mass of solvent in kg. Mass of acetonitrile is solvent (78.6 g or 0.0786 kg). Molality \(m = \frac{0.617}{0.0786} \approx 7.85 \text{ mol/kg}\).
6Step 6: Calculate molarity of methanol
Assuming volumes are additive, total volume \(= 100 + 25 = 125 \text{ mL}\). Convert to liters: \(125 \text{ mL} = 0.125 \text{ L}\). Molarity \(M = \frac{0.617}{0.125} \approx 4.94 \text{ mol/L}\).
Key Concepts
Mole FractionMolalityMolarity
Mole Fraction
The concept of mole fraction is fundamental when discussing solution concentration. It's a way to express the concentration of a component in a mixture. The mole fraction of a substance in a solution is the ratio of the number of moles of that substance to the total number of moles of all substances present. In formula terms, for methanol in this solution, it's given by:
Remember, mole fractions are unitless, making them a versatile tool in quantitative chemistry.
- \( \chi_{\text{CH}_3\text{OH}} = \frac{0.617}{0.617 + 1.914} \approx 0.244 \)
Remember, mole fractions are unitless, making them a versatile tool in quantitative chemistry.
Molality
Molality is another way to express the concentration of a solution but is different from molarity. Molality (\( m \)) is defined as the number of moles of solute per kilogram of solvent. Here, we're calculating the molality of methanol when dissolved in acetonitrile:
Molality provides a reliable measure for scientific calculations especially when dealing with temperature variations.
- \( m = \frac{0.617 \text{ mol}}{0.0786 \text{ kg}} \approx 7.85 \text{ mol/kg} \)
Molality provides a reliable measure for scientific calculations especially when dealing with temperature variations.
Molarity
Molarity (\( M \)) is perhaps the most commonly used unit for concentration in chemistry. It expresses the moles of solute per liter of solution. For methanol in our problem, assuming the solution volumes are additive, the molarity is calculated as:
In laboratory settings, molarity is convenient for preparing solutions and calculating reactant and product volumes in reactions.
- \( M = \frac{0.617 \text{ mol}}{0.125 \text{ L}} \approx 4.94 \text{ mol/L} \)
In laboratory settings, molarity is convenient for preparing solutions and calculating reactant and product volumes in reactions.
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