Problem 151 - A Complete Resource Book in Chemistry for JEE Main

Question

Oleum is considered as a solution of \(\mathrm{SO}_{3}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\), which is obtained by passing \(\mathrm{SO}_{3}\) in concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\). When \(100 \mathrm{~g}\) sample of oleum is diluted with desired weight of \(\mathrm{H}_{2} \mathrm{O}\), then the total mass of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) obtained after dilution is known as \% labeling in oleum. For example, an oleum labeled as '109\% \(\mathrm{H}_{2} \mathrm{SO}_{4}\), means the \(109 \mathrm{~g}\) total mass of pure \(\mathrm{H}_{2} \mathrm{SO}_{4}\) will be formed when \(100 \mathrm{~g}\) of oleum is diluted by \(9 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) which combines with all the free \(\mathrm{SO}_{3}\) present in oleum to form \(\mathrm{H}_{2} \mathrm{SO}_{4}\) as \(\mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}\). \(.9 .0 \mathrm{~g}\) water is added into oleum sample labeled as \(" 112 \% " \mathrm{H}_{2} \mathrm{SO}_{4}\) then the volume of free \(\mathrm{SO}_{3}\) remaining in the solution atl atm pressure and \(0^{\circ} \mathrm{C}\) is (a) \(7.46 \mathrm{~L}\) (b) \(3.73 \mathrm{~L}\) (c) \(11.2 \mathrm{~L}\) (d) \(7.34 \mathrm{~L}\)

Step-by-Step Solution

Answer

The calculations must be revisited for all conditions or components in problem formulation.

Step 1: Understanding the Problem

We have a sample of oleum labeled as '112% H_2SO_4'. This means that when 100 g of the oleum is diluted with 12 g of water (since 112-100=12), it forms 112 g of pure H_2SO_4. It indicates that there is 12 g of SO_3 present in the oleum, which can react with water to form H_2SO_4. We need to find the volume of unreacted SO_3 after adding 9 g of water.

Step 2: Determine Mass of Free SO3

Since the label is 112% H_2SO_4, the oleum initially contains 12 g of SO_3. When 9 g of water is added, 9 g of SO_3 will react according to the reaction SO_3 + H_2O → H_2SO_4. Thus, 3 g of SO_3 remains unreacted because only 9 g of water was added instead of 12 g.

Step 3: Calculate Moles of Free SO3

The molar mass of SO_3 is 80 g/mol (S = 32 g/mol, O = 16 g/mol). The number of moles of unreacted SO_3 is calculated as follows:\[ \text{moles of } SO_3 = \frac{3 \text{ g}}{80 \text{ g/mol}} = 0.0375 \text{ moles} \]

Step 4: Apply Ideal Gas Law

We use the ideal gas law \( PV = nRT \) to find the volume of the free SO_3 gas, where:- \( P = 1 \text{ atm} \)- \( V \) is the volume we need to find- \( n = 0.0375 \text{ moles} \)- \( R = 0.0821 \text{ L atm/mol K} \)- \( T = 273 \text{ K} \) (since 0°C = 273 K)\[ V = \frac{nRT}{P} = \frac{0.0375 \times 0.0821 \times 273}{1} \approx 0.837 \text{ L} \]

Step 5: Conclusion

A mistake occurred in the calculation above, as the number of moles should fit the volume of gas at standard conditions. Re-evaluating, the conditions 1 atm and 0°C imply standard molar volume of the gas is 22.4 L/mol. Re-calculating volume:\[ V = 0.0375 \times 22.4 \approx 0.84 \text{ L} \] actually matches none of the given multiple choice options, so careful calculations and conditions should be revisited.

Step 6: Final Calculation with Volume of SO3

Recalculate by reviewing standard conditions:\[ V = 0.0375 \times 22.4 = 0.84 \text{ L} \]Apologies for incorrect connections to multiple choice. Check consistent conditions across calculations or correct problem conditions for final evaluation -- reevaluate conditions or lab procedure for context adjustments.

Key Concepts

SO3 and H2SO4 Reaction

Oleum is essentially a solution of sulfur trioxide (SO₃) dissolved in sulfuric acid (H₂SO₄). This unique substance is created by bubbling SO₃ through concentrated H₂SO₄. The key feature of oleum is its ability to form additional H₂SO₄ upon dilution.

When water is added to oleum, the SO₃ reacts with the water following the reaction:

• \[ ext{SO}_3 + ext{H}_2 ext{O} ightarrow ext{H}_2 ext{SO}_4 \]

This exothermic reaction results in the conversion of gaseous SO₃ into liquid H₂SO₄, effectively increasing the total mass of sulfuric acid in the solution.

The reverse is also true: when not enough water is available for all the SO₃ to react, some SO₃ remains unreacted in the gaseous form, as shown in the problem where 9 g of water leads only to partial reaction of the available SO₃.

Ideal Gas Law

In scenarios where unreacted gas volumes need to be calculated, like the SO₃ in our oleum exercise, the Ideal Gas Law becomes a powerful tool. The formula is:

• \[ PV = nRT \]

In this equation:

• \(P\) is the pressure (in atm)
• \(V\) is the volume (in L)
• \(n\) is the number of moles of gas
• \(R\) is the universal gas constant (0.0821 L atm/mol K)
• \(T\) is the temperature (in K)

For our specific problem, conditions are set at 1 atm pressure and temperature at 0°C (273 K). By plugging the values into this equation, we can solve for the unknown, which in this case is the volume of the remaining SO₃ gas.

Molar Mass Calculation

To understand the chemical reactions and processes in the exercise, knowing how to calculate molar mass is essential. Molar mass is the mass of one mole of a given substance. It helps convert between mass and moles in stoichiometric calculations.

For sulfur trioxide (SO₃), find the molar mass by summing the atomic masses of sulfur (S) and oxygen (O):

• Sulfur (S) has an atomic mass of approximately 32 g/mol.
• Oxygen (O) has an atomic mass of approximately 16 g/mol.

The molar mass of SO₃ is thus calculated as:

• \[ 32 ext{ g/mol (S)} + 3 imes 16 ext{ g/mol (O)} = 80 ext{ g/mol} \]

Knowing this molar mass enables the conversion of mass of SO₃ into moles for precise calculations in chemical formulas.

Percentage Labeling in Oleum

Percentage labeling in oleum is a representation of how concentrated the solution is in terms of H₂SO₄ formation potential. The label such as "112% H₂SO₄" indicates the amount of pure sulfuric acid that can be produced from diluting 100 g of oleum with a sufficient amount of water. For example:

• A label of "112% H₂SO₄" implies that if 100 g of oleum is mixed with enough water to convert all SO₃ into H₂SO₄, it will yield 112 g of H₂SO₄. Effectively, the label highlights the presence of 12 g of free SO₃ available for the reaction.

Percentage labeling not only assists in determining the concentration but also helps predict the outcome of dilution processes.