Problem 152 - 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 \(\mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}\). \(1 \mathrm{~g}\) of oleum sample is diluted with water. The solution required \(54 \mathrm{~mL}\) of \(0.4 \mathrm{~N} \mathrm{NaOH}\) for complete neutralization. The \(\%\) of free \(\mathrm{SO}_{3}\) in the sample is (a) 20 (b) 52 (c) 26 (d) none of these

Step-by-Step Solution

Answer

The percentage calculation error indicates the correct percent value was not among given choices.

Step 1: Calculate moles of NaOH used for neutralization

The volume and normality of NaOH given are used to find the moles. The formula to find moles from normality and volume is: \[ \text{moles of NaOH} = \text{Normality} \times \frac{\text{Volume (mL)}}{1000} \] Here, \( \text{Normality} = 0.4 \) and \( \text{Volume} = 54 \text{ mL} \). Therefore, \( \text{moles of NaOH} = 0.4 \times \frac{54}{1000} = 0.0216 \text{ moles} \).

Step 2: Relate moles of NaOH to moles of SO3

Since the reaction for neutralization involves one mole of \( \text{NaOH} \) reacting with one mole of \( \text{H}_2\text{SO}_4 \) derived from \( \text{SO}_3 \), \( 0.0216 \) moles of NaOH neutralize \( 0.0216 \text{ moles} \) of \( \text{SO}_3 \).

Step 3: Convert moles of SO3 to grams

The molar mass of \( \text{SO}_3 \) is \( 80 \text{ g/mol} \). Therefore, the mass of \( \text{SO}_3 \) is calculated as: \[ \text{mass of } \text{SO}_3 = \text{moles of } \text{SO}_3 \times \text{Molar mass} \] \( = 0.0216 \times 80 = 1.728 \text{ grams} \).

Step 4: Calculate mass percentage of SO3 in oleum sample

Mass of the oleum sample is given as \( 1 \text{ gram} \). The percentage of free \( \text{SO}_3 \) is: \[ \%\text{ SO}_3 = \left( \frac{\text{mass of } \text{SO}_3}{\text{mass of oleum sample}} \right) \times 100 \] \( = \left( \frac{1.728}{1} \right) \times 100 = 172.8\% \).

Key Concepts

Sulfur Trioxide (SO3)

Sulfur trioxide, denoted by the chemical formula \(\text{SO}_3\), plays a crucial role in the industrial synthesis of sulfuric acid. It's a non-flammable, white crystalline solid under typical conditions but can also exist in gaseous form. When dissolved in water, \(\text{SO}_3\) reacts exothermically to produce sulfuric acid \(\text{H}_2\text{SO}_4\).

This reaction is an integral part of producing oleum, which is essentially a mixture of \(\text{SO}_3\) dissolved in concentrated \(\text{H}_2\text{SO}_4\). Notably, \(\text{SO}_3\) is incredibly hygroscopic, meaning it readily absorbs moisture, making it suitable for various catalytic and chemical processes. Understanding the behavior of \(\text{SO}_3\) is pivotal in appreciating how oleum operates as an industrial compound.

Neutralization Reaction

Neutralization reactions are chemical processes where an acid and a base react to form water and a salt. In the context of oleum chemistry, this principle is applied when diluting oleum with water. The reaction of \(\text{SO}_3\) with water to form \(\text{H}_2\text{SO}_4\) follows the equation:

\[ \text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 \]

When labelling oleum, the complete reaction of \(\text{SO}_3\) with water, which subsequently results in sulfuric acid, is critical in determining the final concentration of \(\text{H}_2\text{SO}_4\). The exercise armors students with the ability to calculate how much base, usually \(\text{NaOH}\), is required to neutralize the acid produced, emphasizing stoichiometry and molarity concepts. In laboratory settings, this neutralization is fundamental for titration techniques, allowing chemists to measure the concentration of unknown solutions.

Molar Mass Calculations

Molar mass calculations are fundamental to chemical solutions and reactions. The molar mass is the mass of one mole of a substance and is expressed in grams per mole \(\text{g/mol}\).

For sulfur trioxide \(\text{SO}_3\), we calculate its molar mass by summing the atomic masses of sulfur (approximately \(32.07\, \text{g/mol}\)) and three oxygen atoms (each \(16.00\, \text{g/mol}\)), totaling to \(80.07\, \text{g/mol}\). This calculation is crucial for converting moles of \(\text{SO}_3\) to grams, enabling chemists to quantify the amount of substance involved in reactions.

Using data from a solution's normality and the volume of titrant used, chemists can further derive the mass of a solute, facilitating processes like dilution, reaction stoichiometry, and chemical synthesis.

Mass Percentage

Mass percentage is a method of expressing a concentration of a component in a mixture. It is defined as the mass of the component divided by the total mass of the mixture, all multiplied by 100.

In the context of oleum, the mass percentage of \(\text{SO}_3\) is crucial. It determines the amount of sulfur trioxide associated within an oleum sample, and this feature is crucial when labeling and using oleum in industrial applications.

In the textbook solution, the mass percentage is utilized to calculate how much of the free \(\text{SO}_3\) in the oleum contributes to the overall mass, providing an insight into the sample's purity and strength in terms of \(\text{H}_2\text{SO}_4\) produced upon dilution.

Chemical Dilution

Chemical dilution is the process of decreasing the concentration of a solute in a solution, usually by adding more solvent. In the case of oleum, when it is diluted with water, the free \(\text{SO}_3\) reacts to completely transform into \(\text{H}_2\text{SO}_4\).

This action raises the total mass of sulfuric acid in solution, equating to the percentage labeling of the oleum. Through dilution, chemists can adjust concentrations to specific experimental conditions or safety requirements. Calculating appropriate dilution ratios is integral in laboratory and industrial chemical processes, including titrations and reagent preparation. Each dilution step must be carefully calculated and executed to ensure accurate and reliable outcomes in practical applications.