Problem 110 - A Complete Resource Book in Chemistry for JEE Main

Question

The partial pressure of oxygen in a flask containin \(16 \mathrm{~g} \mathrm{O}_{2}\) and \(32 \mathrm{~g} \mathrm{SO}_{2}\) is (a) \(1 / 16\) of total pressure (b) \(1 / 2\) of total pressure (c) \(2 / 3\) of total pressure (d) none of the above

Step-by-Step Solution

Answer

The partial pressure of oxygen is \(\frac{1}{2}\) of the total pressure, so the correct answer is (b).

Step 1: Calculate Moles of Each Gas

First, we need to determine the number of moles of each gas in the flask. Use the formula \( ext{moles} = \frac{\text{mass}}{\text{molar mass}} \). For \(\mathrm{O}_2\), the molar mass is \(32\, \mathrm{g/mol}\). Compute: \(\text{moles of }\, \mathrm{O}_2 = \frac{16}{32} = 0.5\, \text{moles}\). For \(\mathrm{SO}_2\), the molar mass is \(64\, \mathrm{g/mol}\). Compute: \(\text{moles of }\, \mathrm{SO}_2 = \frac{32}{64} = 0.5\, \text{moles}\).

Step 2: Find Total Moles

Add the moles of both gases to find the total moles of gas in the flask: \(0.5\, \text{moles of }\, \mathrm{O}_2 + 0.5\, \text{moles of }\, \mathrm{SO}_2 = 1\, \text{mole}\).

Step 3: Calculating the Partial Pressure of O2

The partial pressure of a gas in a mixture is proportional to its mole fraction. The mole fraction of \(\mathrm{O}_2\) is given by \(\frac{\text{moles of } \mathrm{O}_2}{\text{total moles}} = \frac{0.5}{1} = 0.5\). Thus, the partial pressure of \(\mathrm{O}_2\) is 0.5 times the total pressure.

Step 4: Determine Answer Choice

The partial pressure of \(\mathrm{O}_2\) is \(\frac{1}{2}\) of the total pressure (from Step 3). Therefore, the correct answer is option (b): \(\frac{1}{2}\) of the total pressure.

Key Concepts

Mole Concept

Understanding the mole concept is crucial when dealing with chemical quantities and reactions. A mole is simply a unit used to express amounts of a chemical substance. It allows chemists to count entities such as atoms, molecules, or ions in a given substance by relating them to a fixed number known as Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities per mole.
To find the number of moles in a sample, you use the formula: \( ext{moles} = \frac{\text{mass}}{\text{molar mass}} \). Here, the molar mass refers to the mass of one mole of a substance, often found on the periodic table or calculated for compounds by adding up the atomic mass of its constituent elements.
In the context of the exercise, by dividing the mass of each gas by its respective molar mass, we obtain \(0.5\, \text{moles}\) of \(\mathrm{O}_2\) and \(0.5\, \text{moles}\) of \(\mathrm{SO}_2\). By understanding the mole concept, you can see how we convert mass to moles, a foundational skill in chemistry.

Dalton's Law of Partial Pressures

Dalton's Law of Partial Pressures is a key principle in gas chemistry, offering insight into how each gas in a mixture contributes to the total pressure. It states that the total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures exerted by each individual gas in the mixture.
Each gas in the mixture behaves as if it is alone in the container, contributing a partial pressure proportional to its mole fraction. The partial pressure of a gas can be calculated using the formula: \( P_i = x_i \cdot P_{\text{total}} \), where \(x_i\) is the mole fraction of the gas, and \(P_{\text{total}}\) is the total pressure.
In the provided exercise, once we know the mole fraction of oxygen \(\left(0.5\right)\), we can easily find its partial pressure as half of the total pressure, i.e., \(\frac{1}{2}\). This explanation demonstrates how Dalton's Law allows us to break down complex gas mixtures into simpler, manageable parts.

Stoichiometry

Stoichiometry is the section of chemistry that involves calculating the quantities of reactants and products in chemical reactions. It is pivotal to understand the relationships between elements and compounds in a reaction, often utilizing balanced chemical equations.
Stoichiometry allows chemists to predict how much of each substance is needed or produced, using mole relationships and the conversion factors derived from the coefficients in the balanced equation.
Though our exercise does not include a full reaction, understanding stoichiometry still aids in grasping mole relations and transformations. When you determine the moles of gases \(0.5\, \text{moles of each for } \mathrm{O}_2 \text{ and } \mathrm{SO}_2\), it helps link quantities like mass to conceptual ideas such as partial pressures using the mole ratio.
Ultimately, stoichiometry provides the bedrock for more complex calculations involving reactions, further strengthening your chemistry skills and understanding.