Problem 70 - Chemistry and Chemical Reactivity

Question

Vinegar is a \(5 \%\) solution (by weight) of acetic acid in water. Determine the mole fraction and molality of acetic acid. What is the concentration of acetic acid in parts per million (ppm)? Explain why it is not possible to calculate the molarity of this solution from the information provided.

Step-by-Step Solution

Answer

Mole fraction is 0.0156; molality is 0.877 mol/kg; concentration is 50,000 ppm. Molarity can't be determined without solution volume.

Step 1: Understand the Problem

Vinegar is a 5% solution of acetic acid, meaning it contains 5 grams of acetic acid for every 100 grams of solution. We need to find the mole fraction, molality, and parts per million (ppm) of acetic acid. We also need to explain why molarity can't be calculated with given data.

Step 2: Find the Moles of Acetic Acid and Water

Acetic acid (CH₃COOH) has a molar mass of 60.05 g/mol. Since there are 5 grams in the solution, the moles of acetic acid are calculated by dividing the mass by the molar mass: \[ \text{Moles of acetic acid} = \frac{5 \, \text{g}}{60.05 \, \text{g/mol}} \approx 0.0833 \, \text{mol} \]Assume the remaining mass (95g) is water, with water's molar mass being 18.02 g/mol: \[ \text{Moles of water} = \frac{95 \, \text{g}}{18.02 \, \text{g/mol}} \approx 5.274 \, \text{mol} \]

Step 3: Calculate Mole Fraction of Acetic Acid

The mole fraction of acetic acid is calculated by dividing the moles of acetic acid by the total moles of solution (acetic acid + water): \[ \text{Mole fraction} = \frac{0.0833}{0.0833 + 5.274} \approx 0.0156 \]

Step 4: Calculate Molality of Acetic Acid

Molality is the moles of solute per kilogram of solvent. Since we have 0.0833 mol of acetic acid and 0.095 kg of water, the molality is: \[ \text{Molality} = \frac{0.0833 \, \text{mol}}{0.095 \, \text{kg}} \approx 0.877 \, \text{mol/kg} \]

Step 5: Calculate Concentration in Parts Per Million (ppm)

Since ppm is grams of solute per million grams of solution, for 5g of acetic acid in 100g of solution: \[ \text{ppm} = \frac{5 \, \text{g}}{100 \, \text{g}} \times 10^6 = 50000 \, \text{ppm} \]

Step 6: Explain Why Molarity Cannot Be Determined

Molarity involves moles per liter of solution, requiring solution volume. Given data provides mass information only, preventing accurate volume calculation without additional density information.

Key Concepts

Mole Fraction

The concept of mole fraction is crucial in understanding the composition of a solution. It's a way of expressing concentration as a ratio. To calculate the mole fraction of a solute, like acetic acid in vinegar, you divide the number of moles of the solute by the total number of moles present in the solution, including both solute and solvent.
In our example:

• Moles of acetic acid = 0.0833 mol
• Moles of water = 5.274 mol

The total moles in the solution is simply the sum of these two values. The mole fraction is thus:\[\text{Mole fraction of acetic acid} = \frac{0.0833}{0.0833 + 5.274} \approx 0.0156\] This tells us that acetic acid makes up approximately 1.56% of the total mole count in the solution.

Molality

Molality is another important way to express the concentration of a solution. It differs from mole fraction and molarity by the denominator used in its calculation — it's based on the mass, not the volume.
Molality is defined as the moles of solute per kilogram of solvent. In scenarios where temperature varies, molality is often used because it remains constant as opposed to molarity which can change with volume fluctuations.
To find the molality of acetic acid in vinegar:

• We have 0.0833 mol of acetic acid.
• The mass of the water (solvent) is 95 grams or 0.095 kilograms.

The molality calculation then becomes:\[\text{Molality} = \frac{0.0833 \, \text{mol}}{0.095 \, \text{kg}} \approx 0.877 \, \text{mol/kg}\] This indicates there are about 0.877 moles of acetic acid in every kilogram of water in the vinegar solution.

Parts Per Million (ppm)

Parts per million (ppm) is a widely used unit of concentration for very dilute solutions and mixtures. It tells us how many parts of solute exist per million parts of the entire solution.
In our vinegar example, the concentration of acetic acid is given as a percentage, which we can convert into ppm:

• There are 5 grams of acetic acid in 100 grams of solution.
• To convert into parts per million, we make it per million parts.

The conversion appears as follows:\[\text{ppm} = \frac{5 \, \text{g}}{100 \, \text{g}} \times 10^6 = 50000 \, \text{ppm}\]This calculation shows the concentration of acetic acid in vinegar is 50000 ppm, emphasizing its presence within the mixture.

Molar Mass

Molar mass is the mass of a given substance (chemical element or chemical compound) divided by the amount of substance, indicated in moles. It's essential in conversions between grams and moles. The molar mass helps in determining the moles of a substance in a given mass.
Consider acetic acid (CH₃COOH), which has a molar mass:

• Carbon (C), Hydrogen (H), and Oxygen (O) contribute to a molar mass of 60.05 g/mol.

This constant allows us to convert the 5 grams of acetic acid into moles: \[\text{Moles of acetic acid} = \frac{5 \, \text{g}}{60.05 \, \text{g/mol}} \approx 0.0833 \, \text{mol}\]Having this value is vital for calculating other concentration measures like mole fraction and molality, as illustrated in previous sections.