Problem 172 - 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 (c) \(2.28 \, \text{mol} \, \text{kg}^{-1}\).

Step 1: Understand Molarity

The molarity (M) of the solution is the number of moles of solute per liter of solution. In this case, the molarity is given as 2.05 M for acetic acid.

Step 2: Find the Mass of Solution

To find the mass of one liter (1000 mL) of the solution, we multiply the volume by the density of the solution: \(1000 \, \text{mL} \times 1.02 \, \text{g/mL} = 1020 \, \text{g}\).

Step 3: Calculate Moles of Acetic Acid

The moles of acetic acid in 1 liter of solution is given by its molarity: \(2.05 \, \text{mol/L} \times 1 \, \text{L} = 2.05 \, \text{mol}\).

Step 4: Find the Mass of Acetic Acid in Solution

The molar mass of acetic acid (CH₃COOH) is approximately 60.05 g/mol. Thus, the mass of acetic acid in the solution is: \(2.05 \, \text{mol} \times 60.05 \, \text{g/mol} = 123.1025 \, \text{g}\).

Step 5: Calculate Mass of Solvent (Water)

Subtract the mass of acetic acid from the total mass of the solution to find the mass of water: \(1020 \, \text{g} - 123.1025 \, \text{g} = 896.8975 \, \text{g}\). Convert this to kilograms: \(896.8975 \, \text{g} / 1000 = 0.8969 \, \text{kg}\).

Step 6: Calculate Molality

Molality (m) is defined as moles of solute per kilogram of solvent. Using the values calculated: \( \frac{2.05 \text{ mol}}{0.8969 \text{ kg water}} \approx 2.28 \text{ mol/kg}\).

Step 7: Select the Correct Answer

Comparing our calculated molality with the options given, \(2.28 \, \text{mol/kg}\) matches option (c).

Key Concepts

Understanding Molarity

Molarity is a key concept in chemistry, representing the concentration of a solute within a solution. Specifically, it is defined as the number of moles of solute dissolved per liter of solution. This measurement is crucial because it allows chemists to understand how concentrated a solution is, facilitating the preparation of solutions with precise properties.

• Calcuating Molarity: Molarity ( ext{M}) is calculated using the formula: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]


• Example Application: In our exercise, the molarity of acetic acid is given as 2.05 M, indicating that there are 2.05 moles of acetic acid in every liter of solution.

Delving into Molality

Molality is another important way to express the concentration of a solution. Unlike molarity, molality takes into account the mass of the solvent, and is expressed as moles of solute per kilogram of solvent. This measurement helps in scenarios where temperature changes could affect solution volume, making it ideal for precise experiments.

• Calcuating Molality: The formula to determine molality ( ext{m}) is:\[ \text{Molality (m)} = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \]


• Exercise Insight: By calculating the mass of acetic acid and subtracting it from the total solution mass, the solvent mass is derived. This enables the computation of molality, providing a result of 2.28 mol/kg, which best describes the concentration under these conditions.

Density of Solution Explained

The density of a solution provides insight into its mass per unit volume, usually expressed in grams per milliliter (g/mL). This concept is vital when performing calculations that require the mass of a certain volume of solution, such as converting between volume and mass units.

• Calcuating Density: Density (\( ho\)) is determined by:\[ \text{Density} = \frac{\text{mass of solution}}{\text{volume of solution}} \]
  In our scenario, the density is explicitly provided as 1.02 g/mL.


• Using Density in Calculations: With this density value, calculating the total mass of the solution becomes straightforward. For example, for a 1000 mL (or 1 liter) solution, the mass is 1020 g, which is derived by multiplying the volume by the density.

Exploring Acetic Acid

Acetic acid, also known by its chemical formula CH₃COOH, is a common compound with various applications in the laboratory and industry. Understanding its properties is essential when dealing with solutions containing it.

• Properties: Acetic acid is a weak acid and a main component of vinegar, giving it its distinctive smell and sour taste. It has a molar mass of approximately 60.05 g/mol.


• Role in Solutions: In the given exercise, acetic acid acts as the solute. Knowing its molar mass is critical for translating between mass and mole quantities, essential for calculating concentration measures like molarity and molality.