Problem 159 - A Complete Resource Book in Chemistry for JEE Main

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) \(0.44 \mathrm{~mol} \mathrm{~kg}^{-1}\)

Step-by-Step Solution

Answer

The molality of the solution is approximately \(2.28 \mathrm{~mol} \mathrm{~kg}^{-1}\).

Step 1: Understanding Given Data

We have a 2.05 M solution of acetic acid with a density of 1.02 g/mL. This means there are 2.05 moles of acetic acid in 1 liter of solution. Density helps us convert volume to mass.

Step 2: Calculating Mass of Solution

First, we calculate the mass of 1 L (1000 mL) of the solution using the given density: \[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.02 \frac{\text{g}}{\text{mL}} \times 1000 \text{ mL} = 1020 \text{ g} \]

Step 3: Calculating Moles and Mass of Solute

We already know there are 2.05 moles of acetic acid in the solution. The molar mass of acetic acid (\(\text{CH}_3\text{COOH}\)) is about 60.05 g/mol. The mass of acetic acid is: \[ \text{Mass of acetic acid} = 2.05 \text{ moles} \times 60.05 \frac{\text{g}}{\text{mol}} = 123.1025 \text{ g} \]

Step 4: Calculating Mass of Solvent

The mass of the solvent, water, is the total mass of the solution minus the mass of the solute (acetic acid): \[ \text{Mass of solvent (water)} = 1020 \text{ g} - 123.1025 \text{ g} = 896.8975 \text{ g}\]

Step 5: Converting Mass of Solvent to Kilograms

Convert the mass of the water from grams to kilograms as molality is expressed in mol/kg: \[ 896.8975 \text{ g} = 0.8968975 \text{ kg} \]

Step 6: Calculating Molality

Molality is calculated by dividing the moles of solute by the kilograms of solvent: \[ \text{Molality} (m) = \frac{2.05 \text{ moles}}{0.8968975 \text{ kg}} \approx 2.28 \text{ mol/kg} \] This corresponds to option (c).

Key Concepts

Acetic Acid Solution

When dealing with a solution of acetic acid, it's important to understand the components involved. Acetic acid (CH_3 ext{COOH}) is a common component of vinegar, giving it its sour taste and pungent smell. In chemistry, it is often used as a solution in water for various experiments and reactions.
The solution's molarity is expressed as the number of moles of acetic acid per liter of solution. In this context, a 2.05 M solution means that in every liter of solution, there are 2.05 moles of acetic acid.
This concept is crucial because it directly links to the concentration of a solution, which is fundamental when calculating other properties like molality. Understanding the behavior and characteristics of acetic acid solutions allows chemists to predict how they will interact with other substances.

Molality

Molality is one way to measure the concentration of a solution. Unlike molarity, which depends on the total volume of the solution, molality focuses on the amount of solute in a fixed mass of solvent.
The formula used to compute molality is:

• Molality (m) = \( \frac{\text{moles of solute}}{\text{kilograms of solvent}} \)

Molality is particularly useful because it doesn’t change with temperature like molarity does. This is due to the fact that temperature affects volume, but not the mass of the solvent.
For our specific solution, we calculated its molality by determining the number of moles of acetic acid and dividing it by the mass of the solvent in kilograms, which resulted in approximately 2.28 mol/kg. Always remember, the key distinction of molality is its relation to solvent mass, not solution volume.

Density

Density is a property that describes how much mass there is in a given volume of a substance. It’s expressed as the ratio of mass to volume. For our solution, the density was given as 1.02 g/mL.
This concept is vital because it allows us to convert between the volume of the solution and its mass. By knowing the density and the volume of the solution, we can easily find out its total mass.
Given that the density of the acetic acid solution is 1.02 g/mL, we could calculate that 1 liter (or 1000 mL) of this solution has a mass of 1020 grams. This information is critical for the subsequent calculation of solvent quantity and molality.

Moles of Solute

The concept of "moles of solute" refers to the amount of a substance present in a solution, expressed in moles. It is an essential measure as it helps to determine how concentrated a solution is.
In our scenario, there are 2.05 moles of acetic acid dissolved in the solution, which is the key information used to calculate both the molality and the amount of solute in grams.
To find the actual mass of acetic acid, we use the molar mass of acetic acid, which is approximately 60.05 g/mol. By multiplying the moles by the molar mass, we obtain the mass of acetic acid as 123.1025 g.
This calculation allows us to differentiate the solute from the solvent, and further analyze the properties of the solution with precision. Understanding the number of moles and molar mass is essential for any chemical preparation or analysis.