Question

A solution is made containing \(14.6 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{OH}\) in \(184 \mathrm{~g}\) \(\mathrm{H}_{2} \mathrm{O}\). Calculate (a) the mole fraction of \(\mathrm{CH}_{3} \mathrm{OH}\), (b) the mass percent of \(\mathrm{CH}_{3} \mathrm{OH}\), (c) the molality of \(\mathrm{CH}_{3} \mathrm{OH}\).

Short answer

In conclusion: a) The mole fraction of CH₃OH = 0.0457 b) The mass percent of CH₃OH = 7.35% c) The molality of CH₃OH = 2.48 mol/kg.

Step-by-step solution

Calculate moles of CH₃OH and H₂O

First, we need to determine the number of moles of methanol and water in the solution. To do this, we will use their respective molar masses: For CH₃OH, molar mass = 12.01 (C) + 3 x 1.01 (H) + 1.01 (H) + 16.00 (O) = 32.05 g/mol For H₂O, molar mass = 2 x 1.01 (H) + 16.00 (O) = 18.02 g/mol. Now, we can find the moles of each substance: Moles of CH₃OH = mass / molar mass = 14.6 g / 32.05 g/mol = 0.4556 mol Moles of H₂O = mass / molar mass = 184 g / 18.02 g/mol = 10.22 mol.

Compute the mole fraction of CH₃OH

The mole fraction (X) is the ratio of the number of moles of a component to the total number of moles in the solution. In this case, we want to find the mole fraction of methanol: X(CH₃OH) = moles of CH₃OH / (moles of CH₃OH + moles of H₂O) = 0.4556 mol / (0.4556 mol + 10.22 mol) = 0.0457.

Determine the mass percent of CH₃OH

The mass percent (wt%) is the ratio of the mass of a component to the total mass of the solution multiplied by 100. Here, we will calculate the mass percent of methanol: wt%(CH₃OH) = (mass of CH₃OH / total mass) x 100 = (14.6 g / (14.6 g + 184 g)) x 100 = (14.6 g / 198.6 g) x 100 = 7.35%.

Calculate the molality of the solution

Molality (m) is the number of moles of solute (in our case, CH₃OH) per kilogram of solvent (in our case, H₂O). Now, we can find the molality of the solution: m(CH₃OH) = moles of CH₃OH / mass of H₂O in kg = 0.4556 mol / (184 g / 1000) = 0.4556 mol / 0.184 kg = 2.48 mol/kg. In conclusion, we have found: a) The mole fraction of CH₃OH = 0.0457 b) The mass percent of CH₃OH = 7.35% c) The molality of CH₃OH = 2.48 mol/kg.

Key concepts

Mole Fraction

The mole fraction is a way to express the concentration of a component in a solution. It's calculated by dividing the number of moles of a particular substance by the total number of moles in the solution. In this context, we're interested in the mole fraction of methanol (CH₃OH).
In our example, we first calculated the moles of methanol, which came out to be 0.4556 moles, and the moles of water, which were 10.22 moles. To find the mole fraction, we used the formula: \[ X(\text{CH}_3\text{OH}) = \frac{\text{moles of CH}_3\text{OH}}{\text{moles of CH}_3\text{OH} + \text{moles of H}_2\text{O}} = \frac{0.4556}{0.4556 + 10.22} \approx 0.0457 \]This means that methanol makes up about 4.57% of the total mole count in the solution. Mole fractions are dimensionless numbers and provide insight into the proportion of one component in a solution relative to the entire mixture.

Mass Percent

Mass percent gives us a straightforward understanding of the concentration by showing how much of the solution's mass comes from a specific component. It's calculated with the formula: \[ \text{Mass Percent} = \left( \frac{\text{mass of the component}}{\text{total mass of the solution}} \right) \times 100 \]In our exercise, the mass of methanol was 14.6 grams while the total mass of solution was 198.6 grams (14.6 g + 184 g). The mass percent calculation for methanol looked like this: \[ \text{wt% (CH}_3\text{OH)} = \left( \frac{14.6}{198.6} \right) \times 100 \approx 7.35\% \]This result tells us that 7.35% of our total solution's weight is due to methanol. Mass percent allows us to express the concentration in a tangible way, using easily understood units of measurement, namely grams.

Molality

Molality is yet another valuable measure of concentration, which is particularly useful when dealing with temperature-sensitive calculations. Molality measures the moles of solute per kilogram of solvent—not solution. To derive molality, we used the methanol's moles and the water's weight in kilograms:\[ \text{molality (m)} = \frac{\text{moles of CH}_3\text{OH}}{\text{mass of H}_2\text{O in kg}} = \frac{0.4556}{0.184} = 2.48 \, \text{mol/kg} \]In our scenario, we took the moles of methanol, which were 0.4556 moles, and divided them by the mass of water converted into kilograms (184 g = 0.184 kg).
Thus, the solution’s molality is 2.48 mol/kg, showing the concentration of methanol based purely on the mass of the water rather than the solution. This avoids the dependency on volume, which can change with temperature fluctuations.