Problem 48 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

\(8.0575 \times 10^{-2} \mathrm{~kg}\) of Glauber's salt is dissolved in water to obtain 1 \(\mathrm{dm}^{3}\) of a solution of density \(1077.2 \mathrm{~kg} \mathrm{~m}^{3}\). Calculate the molarity, molality and mole fraction of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in the solution. [1994 - 3 Marks]

Step-by-Step Solution

Answer

Molarity: 0.025 M, Molality: 0.0234 mol/kg, Mole fraction: 4.2 x 10^-4.

Step 1: Convert Mass to Moles of Solute

First, calculate the moles of Glauber's salt \( \text{Na}_2\text{SO}_4\cdot10\text{H}_2\text{O} \). The molar mass is \(2(23) + 32 + 4(16) + 10(18) = 322 \ \text{g/mol}\). Convert the 8.0575 g to kg: \(8.0575 \text{ kg} \times 10^{-2} = 8.0575 \text{ g}\). Now, moles of solute = \(\frac{8.0575}{322} \approx 0.025 \text{ mol}\).

Step 2: Calculate Molarity

Molarity \(M\) is defined as the number of moles of solute per liter of solution. Given the volume of solution is 1 \(\text{dm}^3\) which is 1 liter, the molarity is \(\frac{0.025 \text{ mol}}{1 \text{ liter}} = 0.025 \text{ M}\).

Step 3: Calculate Mass of Solvent

To calculate the mass of solvent, first find the total mass of the solution using density: \(\text{mass} = \text{density} \times \text{volume} = 1077.2 \text{ kg/m}^3 \times 1 \text{ dm}^3 \times 10^{-3} \text{ m}^3/\text{dm}^3 = 1.0772 \text{ kg}\). Subtract the mass of solute to get the mass of the solvent: \(1.0772 \text{ kg} - 0.0080575 \text{ kg} = 1.0691425 \text{ kg}\).

Step 4: Calculate Molality

Molality \(m\) is the number of moles of solute per kilogram of solvent. With 0.025 moles of solute and 1.0691425 kg of solvent, molality is \(\frac{0.025}{1.0691425} \approx 0.0234 \text{ mol/kg}\).

Step 5: Calculate Mole Fraction of \(\text{Na}_2\text{SO}_4\)

Mole fraction \(X\) is the number of moles of solute divided by the total number of moles (solute plus solvent). First, calculate moles of water (solvent): \(\frac{1069.1425 \text{ g}}{18 \text{ g/mol}} \approx 59.397 \text{ mol}\). \(X_{\text{Na}_2\text{SO}_4} = \frac{0.025}{0.025 + 59.397} \approx 4.2 \times 10^{-4}\).

Key Concepts

Glauber's salt in Chemical Analysis

Glauber's salt, formally known as sodium sulfate decahydrate (\( ext{Na}_2 ext{SO}_4 ext{·}10 ext{H}_2 ext{O}\)), is an intriguing compound used in chemistry for various applications, including in solution preparation. When working with Glauber's salt, it is important to remember its hydrated form. This means that the compound incorporates water molecules within its crystal structure. In this case, there are 10 water molecules for every formula unit of sodium sulfate, which makes up nearly 56% of the total molar mass. When carrying out molarity calculations, like in the given exercise, the first step involves finding the moles of the compound based on its molar mass. The molar mass accounts for both the sodium sulfate and the water in its hydrated form. For Glauber's salt, this is 322 \( ext{g/mol}\), which includes the mass of sodium ions, sulfate ions, and the associated water molecules. Being aware of these details is critical when performing density calculations and stoichiometric computations based on this substance.

Understanding Density Calculations

Density is a fundamental concept in chemistry, helping us relate the mass of a substance to its volume. In the problem, we need to use the density of the solution to find its total mass. We know the volume of the solution is 1 \( ext{dm}^3\) or 1 liter, and its density is 1077.2 \( ext{kg/m}^3\). By multiplying these values, we derive the total mass of the solution.

• The formula is: \[ ext{density} = \frac{ ext{mass}}{ ext{volume}}\]
• For this problem: mass = density \( imes\) volume = 1077.2 \( ext{kg/m}^3\) \( imes\) 1 \( ext{dm}^3\), noting that 1 \( ext{dm}^3\) equals 0.001 \( ext{m}^3\) in calculations.

After obtaining the total mass of the solution, subtract the mass of the solute (Glauber's salt) to determine the mass of only the solvent, i.e., water. This approach showcases how density plays a crucial role in both solution concentration and chemical stoichiometry, making it a vital part of any molarity or molality computation.

Understanding Solution Concentration

Solution concentration is a measure of how much solute is dissolved in a solvent. Molarity (\(M\)) and molality (\(m\)) are two central parameters that provide insights into concentration.

• Molarity (\(M\)): This measures the number of moles of solute per liter of solution. For the task at hand, the molarity was calculated as 0.025 \( ext{M}\) from the 0.025 moles of Glauber's salt dissolved in 1 \( ext{dm}^3\) of solution.
• Molality (\(m\)): Unlike molarity, molality measures the moles of solute per kilogram of solvent. The task provided a value of 0.0234 \( ext{mol/kg}\) calculated based on the kilograms of the water solvent.

By understanding these concepts and calculations, students can determine how concentrated a solution is, which can be crucial for quantitative chemical analysis and reactions involving stoichiometry.

Mastering Chemical Stoichiometry

Chemical stoichiometry involves calculations relating to the quantities of reactants and products in chemical reactions. Here, it allows us to compute how many moles of a substance are present, including complications like hydrates. Calculating the mole fraction is a key part of stoichiometry. The mole fraction tells us the proportion of the solute in relation to the entire solution mixture—involving both solute and solvent.In our exercise:

• The mole fraction of \( ext{Na}_2 ext{SO}_4\) is calculated as the number of moles of Glauber's salt divided by the total moles (solute + water).
• Using the formula: \[ ext{Mole Fraction}, \,X\ = \frac{n_{ ext{solute}}}{n_{ ext{solute}} + n_{ ext{solvent}}} \]

For Glauber's salt in the solution, the mole fraction comes out to approximately 4.2 \( imes\) 10^{-4}. This low value reflects the relatively small amount of solute in comparison to a large amount of water, a common scenario in dilute solutions. Understanding these calculations helps in grasping the foundational aspects of chemical reactions and solution properties.