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 (c) \(2.28 \mathrm{~mol/kg}\).

Step 1: Understand the Given Information

We have a solution with a molarity of \(2.05 \mathrm{M}\) and a density of \(1.02 \mathrm{g/mL}\). We need to find the molality and compare it with the given options.

Step 2: Convert Molarity to Moles in 1 L Solution

Molarity \( (M)\) is moles of solute per liter of solution. For \(2.05 \mathrm{M acetic}\) acid, it means there are \(2.05\) moles of acetic acid in \(1\) liter of solution.

Step 3: Calculate Mass of Solution in grams

The volume of the solution is \(1000 \mathrm{~mL}\) (since 1 L = 1000 mL). Given the density \( = 1.02 \mathrm{~g/mL}\), the mass of \(1000 \mathrm{~mL}\) solution is: \[ \text{Mass of solution} = 1.02 \times 1000 = 1020 \mathrm{~g} \]

Step 4: Find Mass of Acetic Acid

The molar mass of acetic acid \((\mathrm{CH}_3\mathrm{COOH})\) is approximately \(60 \mathrm{~g/mol}\). Hence, the mass of \(2.05\) moles of acetic acid is: \[ \text{Mass of acetic acid} = 2.05 \times 60 = 123 \mathrm{~g} \]

Step 5: Calculate Mass of Solvent (Water)

Subtract the mass of acetic acid from the total mass of the solution to get the mass of the solvent (water): \[ \text{Mass of water} = 1020 \mathrm{~g} - 123 \mathrm{~g} = 897 \mathrm{~g} \] Convert this mass to kilograms: \[ 897 \mathrm{~g} = 0.897 \mathrm{~kg} \]

Step 6: Calculate Molality

Molality \((m)\) is moles of solute per kilogram of solvent. Using the mass of water (solvent):\[ \text{Molality} = \frac{2.05 \mathrm{~mol}}{0.897 \mathrm{~kg}} = 2.28 \mathrm{~mol/kg} \]

Step 7: Verify the Answer

Check the calculated molality against the provided options: The molality \(2.28 \mathrm{~mol/kg}\) matches option (c).

Key Concepts

Molarity to Molality Conversion

Molarity and molality are two ways to express the concentration of a solution, but they are not interchangeable without some calculations. Molarity \(M\) is defined as the number of moles of solute per liter of solution. In contrast, molality \(m\) is the number of moles of solute per kilogram of solvent. To convert between these two:

• First, identify the molarity of the solution. For example, a \(2.05 \, M\) acetic acid solution means there are \(2.05\) moles of acetic acid in \(1\) liter of solution.
• Next, calculate the total mass of the \(1\) liter solution using its density. This will allow you to find the mass of the solvent by subtracting the mass of the acetic acid.
• Finally, divide the moles of solute by the mass of the solvent (in kilograms) to find the molality.

The essence of converting molarity to molality lies in understanding the difference in how each measures concentration, requiring knowledge of both the solution's volume and the solvent's mass.

Density of Solutions

Density is a critical factor in converting molarity to molality. It tells us how much mass there is per unit volume of a solution. For the \(2.05 \, M\) acetic acid solution in question, the density is \(1.02 \, \text{g/mL}\). This information helps us calculate the total mass of our solution:

• Consider the volume of the solution, which is \(1000 \, \text{mL}\) or \(1\text{L}\).
• By multiplying the density by the volume, we get the total mass of the solution. For instance: \[ \text{Mass of solution} = 1.02 \, \text{g/mL} \times 1000 \, \text{mL} = 1020 \, \text{g} \]

This calculated mass is essential as it provides a basis for further calculations, primarily determining the solvent's mass after accounting for the solute's mass. Understanding the density is crucial for making these conversions precise and meaningful.

Mass Percent and Mass of Solvent

Mass percent is a way of expressing the concentration of a component in a mixture. In our case, once we know the total mass of the solution (1020 g) and the mass of acetic acid added (123 g), we can determine the mass of water:

• Subtract the mass of acetic acid (the solute) from the total mass of the solution to find the solvent's mass, which is water.
• Use this subtraction to ensure you're only accounting for the water when calculating molality: \[ \text{Mass of water} = 1020 \, \text{g} - 123 \, \text{g} = 897 \, \text{g} = 0.897 \, \text{kg} \]

Understanding mass percent and being able to isolate the mass of the solvent is a crucial step. It allows you to focus solely on the solvent when calculating concentrations like molality, emphasizing the distinction between the solution's and solvent's contributions to concentration metrics.

Molar Mass Calculations

Molar mass is the mass of one mole of a substance. When dealing with acetic acid \(\text{CH}_3\text{COOH}\), for instance, it's crucial to correctly calculate its molar mass to find out how much mass corresponds to the moles present in the solution:

• The molar mass of acetic acid is about \(60 \, \text{g/mol}\).
• Knowing the number of moles of acetic acid, multiply by the molar mass to find its total mass: \[ \text{Mass of acetic acid} = 2.05 \, \text{mol} \times 60 \, \text{g/mol} = 123 \, \text{g} \]

Accurate molar mass calculations are foundational in converting between different concentrations since they provide the precise mass of the solute needed for other calculations. Being precise and confident with molar mass ensures that you can reliably proceed with concentration conversions.