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Chapter 13: Problem 49

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?

Short Answer

Expert verified
Mole fraction: 0.244. Molality: 7.85 mol/kg. Molarity: 4.94 mol/L.

Step by step solution

01

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, \( \text{mass} = \text{density} \times \text{volume} = 0.791 \times 25.0 = 19.775 \text{ g}\).
02

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, \( \text{mass} = \text{density} \times \text{volume} = 0.786 \times 100 = 78.6 \text{ g}\).
03

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 \( = \frac{78.6}{41.05} \approx 1.914 \text{ mol}\).
04

Calculate mole fraction of methanol

Mole fraction of methanol \(\chi_{\text{CH}_3\text{OH}}\) is calculated using the formula: \(\chi_{\text{CH}_3\text{OH}} = \frac{\text{moles of methanol}}{\text{moles of methanol} + \text{moles of acetonitrile}} = \frac{0.617}{0.617 + 1.914} \approx 0.244\).
05

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 \(m = \frac{0.617}{0.0786} \approx 7.85 \text{ mol/kg}\).
06

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 \(M = \frac{0.617}{0.125} \approx 4.94 \text{ mol/L}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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:
  • \( \chi_{\text{CH}_3\text{OH}} = \frac{0.617}{0.617 + 1.914} \approx 0.244 \)
The total moles in the solution here include both methanol and acetonitrile. The sum of the mole fractions of all components in a solution always equals one. This concept is especially useful when dealing with vapor pressures and is often used in chemical engineering and thermodynamics.
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:
  • \( m = \frac{0.617 \text{ mol}}{0.0786 \text{ kg}} \approx 7.85 \text{ mol/kg} \)
A significant advantage of molality is its independence from temperature, because it is based on mass rather than volume, which can change with temperature. This makes molality particularly useful in scenarios involving temperature changes, like in studying colligative properties (boiling point elevation and freezing point depression).
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:
  • \( M = \frac{0.617 \text{ mol}}{0.125 \text{ L}} \approx 4.94 \text{ mol/L} \)
Molarity is temperature-dependent because the volume of solutions generally changes with temperature. This is an important consideration when dealing with precise chemical reactions. When you measure molarity, it's also crucial to know the total volume, including both solute and solvent.
In laboratory settings, molarity is convenient for preparing solutions and calculating reactant and product volumes in reactions.

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Most popular questions from this chapter

Would you expect alanine (an amino acid) to be more soluble in water or in hexane?

If you compare the solubilities of the noble gases in water, you find that solubility increases from smallest atomic weight to largest, \(\mathrm{Ar}<\mathrm{Kr}<\mathrm{Xe}\). Which of the following statements is the best explanation? (a) The heavier the gas, the more it sinks to the bottom of the water and leaves room for more gas molecules at the top of the water. (b) The heavier the gas, the more dispersion forces it has, and therefore the more attractive interactions it has with water molecules. (c) The heavier the gas, the more likely it is to hydrogenbond with water. (d) The heavier the gas, the more likely it is to make a saturated solution in water.

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The Baltic Sea has a salinity of \(1.0 \%\), that is, its water contains \(10 \mathrm{~g}\) of salt for every liter of solution. Assuming that the solute consists entirely of \(\mathrm{NaCl}\) (in fact, over \(90 \%\) of the salt is indeed \(\mathrm{NaCl}\), calculate the osmotic pressure of this seawater at \(15^{\circ} \mathrm{C}\).
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