Problem 49 - Chemistry: The Central Science in SI Units, Global Edition

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

Answer

In conclusion, the mole fraction of methanol in the solution is 0.244, the molality of the solution is 7.84 mol/kg, and the molarity of methanol in the solution is 4.94 mol/L, assuming that the volumes are additive.

Step 1: Find the mass of acetonitrile and methanol

First, we need to find the mass of each substance in the solution. To do this, we will use the densities and volumes of each substance. Mass of acetonitrile (CH₃CN): Density = 0.786 g/mL Volume = 100 mL Mass = Density × Volume \(Mass_{CH_3CN} = 0.786 \ g/mL \times 100 \ mL = 78.6 \ g\) Mass of methanol (CH₃OH): Density = 0.791 g/mL Volume = 25 mL Mass = Density × Volume \(Mass_{CH_3OH} = 0.791 \ g/mL \times 25 \ mL = 19.775 \ g\)

Step 2: Convert mass to moles

To calculate the mole fraction and molality, we need to know the number of moles of each substance. To do this, we will use the molar mass of each substance to convert the mass to moles. Molar mass of acetonitrile (CH₃CN) = 12.01 g/mol (C) + 3 × 1.01 g/mol (H) + 14.01 g/mol (N) = 41.05 g/mol Molar mass of methanol (CH₃OH) = 12.01 g/mol (C) + 4 × 1.01 g/mol (H) + 16.00 g/mol (O) = 32.04 g/mol Number of moles of acetonitrile: \(n_{CH_3CN} = \frac{Mass_{CH_3CN}}{Molar \ Mass_{CH_3CN}}\) \(n_{CH_3CN} = \frac{78.6 \ g}{41.05 \ g/mol} = 1.914 \ mol\) Number of moles of methanol: \(n_{CH_3OH} = \frac{Mass_{CH_3OH}}{Molar \ Mass_{CH_3OH}}\) \(n_{CH_3OH} = \frac{19.775 \ g}{32.04 \ g/mol} = 0.617 \ mol\)

Step 3: Calculate the mole fraction of methanol

To find the mole fraction of methanol in the solution, we can use the following equation: Mole fraction of methanol (Xₘ) = \(\frac{n_{CH_3OH}}{n_{CH_3OH} + n_{CH_3CN}}\) \(X_m = \frac{0.617 \ mol}{0.617 \ mol + 1.914 \ mol} = 0.244\) The mole fraction of methanol in the solution is 0.244.

Step 4: Calculate the molality of the solution

To find the molality of the solution, we can use the following equation: Molality (m) = \(\frac{n_{CH_3OH}}{Mass_{CH_3CN} (kg)}\) \(m = \frac{0.617 \ mol}{0.0786 \ kg} = 7.84 \ mol/kg\) The molality of the solution is 7.84 mol/kg.

Step 5: Calculate the molarity of methanol in the solution

Assuming that the volumes are additive, we know that the total volume of the solution is 100 mL (CH₃CN) + 25 mL (CH₃OH) = 125 mL or 0.125 L. To find the molarity of methanol in the solution, we can use the following equation: Molarity (M) = \(\frac{n_{CH_3OH}}{Volume_{solution} (L)}\) \(M = \frac{0.617 \ mol}{0.125 \ L} = 4.94 \ mol/L\) The molarity of methanol in the solution is 4.94 mol/L. In conclusion, the mole fraction of methanol in the solution is 0.244, the molality of the solution is 7.84 mol/kg, and the molarity of methanol in the solution is 4.94 mol/L, assuming that the volumes are additive.

Key Concepts

Density

Density is a key property of matter that explains how much mass is contained within a specific volume. It is expressed as mass per unit volume, especially in grams per milliliter for liquids. For example, in our solution, the density of acetonitrile is 0.786 g/mL and the density of methanol is 0.791 g/mL.

When you need to find the mass of a liquid, you can easily do so using its density. just multiply the density by the volume of the liquid. For instance, for 100 mL of acetonitrile, the mass is calculated by 0.786 g/mL × 100 mL, resulting in 78.6 grams. Likewise, for 25 mL of methanol, it would be 0.791 g/mL × 25 mL, giving 19.775 grams.

Mole Fraction

The mole fraction is a way to express the concentration of a particular component in a solution. It represents the ratio of moles of that component to the total moles of all components in the solution. This is particularly useful in chemical processes where the relative proportions of different substances are significant.

For example, to find the mole fraction of methanol in our solution:

• Determine the number of moles of methanol using its mass and molar mass, 0.617 moles.
• Find the total moles by adding the moles of both acetonitrile and methanol, 0.617 + 1.914 = 2.531 moles.

The mole fraction of methanol is then calculated as the moles of methanol divided by the total moles, resulting in a mole fraction of 0.244.

Molarity

Molarity is a common measure of solution concentration in chemistry. It is defined as the number of moles of solute (in this case, methanol) per liter of solution. This measure is particularly valuable because it directly relates to volume, making it very practical for laboratory settings where reactions are volume-based.

Using our example, we assume additive volumes. The total volume of our solution is 125 mL or 0.125 L. To find the molarity:

• Use the number of moles of methanol, which is 0.617 mol.
• Divide by the solution volume in liters, 0.125 L.

This results in a molarity of 4.94 mol/L for methanol.

Molality

Molality is another way to express solution concentration. It is defined as moles of solute per kilogram of solvent, not solution. A key advantage of molality is that it does not change with temperature, unlike molarity.

To calculate the molality of methanol in the solution, use:

• Moles of methanol: 0.617 moles.
• Mass of the solvent (acetonitrile) in kg: 78.6 g = 0.0786 kg.

The molality can hence be calculated by dividing 0.617 moles by 0.0786 kg, yielding a molality of 7.84 mol/kg. This shows how many moles of methanol are present per kilogram of the acetonitrile solvent.