Question

Density of a \(2.05 \mathrm{M}\) solution of acetic acid in water is \(1.02 \mathrm{~g} / \mathrm{ml}\). The molality of the solution is a. \(1.14 \mathrm{~mol} \mathrm{~kg}^{-1}\) b. \(3.28 \mathrm{~mol} \mathrm{~kg}^{-1}\) c. \(2.28 \mathrm{~mol} \mathrm{~kg}^{-1}\) d. \(0.44 \mathrm{~mol} \mathrm{~kg}^{-1}\)

Short answer

The molality of the solution is 2.28 mol kg^{-1} (option c).

Step-by-step solution

Understand the Problem

We are given a molarity of acetic acid solution and the density of the solution. We need to find the molality of the solution. Recall that molarity (M) is defined as moles of solute per liter of solution, while molality (m) is moles of solute per kilogram of solvent.

Determine Moles of Solute

The molarity of the solution is given as 2.05 M, which means 2.05 moles of acetic acid are present in 1 liter (1000 ml) of solution.

Use Density to Find Total Mass of the Solution

The density of the solution is 1.02 g/ml. Therefore, the mass of 1000 ml of solution is:\[ \text{Mass of solution} = 1.02 imes 1000 = 1020 \text{ grams} \]

Calculate Mass of Solvent (Water)

We know the total mass of the solution, and we need to subtract the mass of the solute (acetic acid). The molar mass of acetic acid (\(\text{CH}_3\text{COOH}\)) is approximately 60.05 g/mol. Thus, the mass of 2.05 moles of acetic acid is:\[ \text{Mass of solute (acetic acid)} = 2.05 \times 60.05 = 123.1025 \text{ grams} \]The mass of the solvent (water) is:\[ \text{Mass of solvent} = \text{Total mass} - \text{Mass of solute} = 1020 - 123.1025 = 896.8975 \text{ grams} \]

Calculate Molality

Convert the mass of the solvent to kilograms:\[ \text{Mass of solvent in kg} = \frac{896.8975}{1000} = 0.8968975 \text{ kg} \]Molality (m) is calculated as moles of solute per kg of solvent:\[ \text{Molality} = \frac{2.05}{0.8968975} = 2.285 \text{ mol/kg} \]

Choose the Closest Answer

The calculated molality is 2.285 mol/kg, which rounds to 2.28 mol/kg. The closest answer from the given options is: **c. 2.28 mol kg^{-1}**

Key concepts

Density of Solution

Understanding the concept of density is crucial when dealing with solutions. Density is defined as mass per unit volume of a substance. It helps us determine how much the solution weighs based on its volume.
To calculate the mass of a solution, you simply multiply its density by its volume. For example, given a density of \(1.02 \text{ g/ml}\) and a volume of \(1000\text{ ml}\), the total mass of the solution is \(1.02 \times 1000 = 1020 \text{ grams}\).
This simple relationship between mass, density, and volume can be applied in various contexts. It's especially useful for calculating other properties of solutions, like molality, when the density is known.

• Density provides a bridge between the physical and chemical properties of a solution.
• It allows for conversion between volume-based and mass-based measurements.
• Understanding density is key to unlocking more complex solution calculations.

Molarity to Molality Conversion

Converting between molarity and molality can be initially confusing, but it becomes straightforward with practice. Molarity is the number of moles of solute per liter of solution, whereas molality is the number of moles of solute per kilogram of solvent. Knowing this distinction is vital for proper conversion.
Given the molarity (\(2.05 \text{ M}\)) and the density of the solution, we start by calculating the total mass of the solution. Next, subtract the mass of the solute to find the mass of the solvent. Then convert this mass from grams to kilograms.
Finally, calculate the molality by dividing the number of moles of solute by the mass of the solvent in kilograms:
\[\text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}\]
This calculation strikes at the heart of understanding how solute concentration metrics vary between volume-based and mass-based measures. These conversions are fundamental tools in chemistry, biology, and many applied sciences.

• Molarity is volume-based while molality is mass-based.
• The conversion requires careful mass calculations, including the use of density.
• The distinction aids in research and industry applications where conditions may vary.

Mass of Solute Calculation

Calculating the mass of a solute within a solution is essential to understand its composition. For instance, to determine how much solute (such as acetic acid) is present, you multiply the moles of solute by its molar mass.
In this example, \(2.05\) moles of acetic acid (with a molar mass of \(60.05\text{ g/mol}\)) results in:
\[\text{Mass of solute} = 2.05 \times 60.05 = 123.1025 \text{ grams}\]
This information is paramount in calculating the solvent’s mass and, subsequently, the molality.
Understanding these calculations allows you to draw insights about solute concentration and solution behavior, such as boiling and freezing points, which are influenced by the solute's precise mass.

• The mass of the solute is determined by the product of moles and molar mass.
• Knowing the solute's mass enables the calculation of other solution properties.
• It is critical for precise solution preparation and analysis.