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 total pressure, so the correct answer is (b).

Step 1: Calculate the number of moles of each gas

First, we need to find the number of moles of each gas. The molar mass of \( \mathrm{O}_2 \) is 32 g/mol and the molar mass of \( \mathrm{SO}_2 \) is 64 g/mol. Thus, the moles of \( \mathrm{O}_2 = \frac{16 \text{ g}}{32 \text{ g/mol}} = 0.5 \) moles, and the moles of \( \mathrm{SO}_2 = \frac{32 \text{ g}}{64 \text{ g/mol}} = 0.5 \) moles.

Step 2: Calculate the mole fraction of oxygen

The total moles of gas \(= 0.5 \text{ moles } \mathrm{O}_2 + 0.5 \text{ moles } \mathrm{SO}_2 = 1.0 \text{ moles}.\) The mole fraction of \( \mathrm{O}_2 \) is \(\frac{\text{moles of } \mathrm{O}_2}{\text{total moles}} = \frac{0.5}{1.0} = 0.5.\)

Step 3: Determine the partial pressure of oxygen

The partial pressure of a gas is equal to its mole fraction times the total pressure. Hence, the partial pressure of \( \mathrm{O}_2 \) is \(0.5 \times \text{total pressure} = \frac{1}{2} \text{ of total pressure}.\)

Key Concepts

Mole Fraction

The concept of mole fraction is a key part in understanding gas mixtures and partial pressures. In simple terms, the mole fraction of a component in a mixture is the ratio of the number of moles of that component to the total number of moles of all components. For example:

\[X_{O_2} = \frac{\text{Moles of } O_2}{\text{Total moles in mixture}}\]

In our problem, we calculate the mole fraction of oxygen as \(0.5\). This means that the oxygen gas makes up 50% of the molecules present in the flask. It is important to understand that the mole fraction is a dimensionless number, typically expressed as a decimal between 0 and 1.

Using mole fraction allows us to understand the proportion of gases in a mixture, which is crucial for further calculations such as determining partial pressure.

Molar Mass Calculation

Calculating the molar mass is an essential step in stoichiometry, especially when working with chemical reactions and gas mixtures. The molar mass is the mass of one mole of a substance, which is typically expressed in grams per mole (g/mol).

To find the molar mass, sum the atomic masses of all atoms in a molecule. For instance:

• \(\text{For } O_2: 16 \, \text{g/mol per O atom} \times 2 = 32 \, \text{g/mol}\)
• \(\text{For } SO_2: 32 \, \text{g/mol per S atom} + (16 \, \text{g/mol per O atom} \times 2) = 64 \, \text{g/mol}\)

Calculating molar mass allows you to convert between grams and moles, a common necessity in chemical calculations. Understanding how to determine molar mass ensures you can accurately measure out materials and predict reaction yields.

Gas Mixtures

Gas mixtures contain multiple types of gases within a space. Each gas in a mixture behaves independently according to the ideal gas law, allowing us to calculate individual partial pressures. The combined pressures of all gases contribute to the total pressure.

Individual gas properties, such as mole fractions, influence their proportion in the mixture. For example, if oxygen and sulfur dioxide are in a flask, with \(0.5\) mole fraction each, and considering ideal behavior, each gas contributes equally to the mixture's properties.

Understanding gas mixtures and how different gases interact allows you to predict behaviors like diffusion, reaction rates, and pressure equilibrium, all of which are critical for practical applications.

Stoichiometry

Stoichiometry deals with the quantitative relationships between reactants and products in chemical reactions. It is crucial in predicting how much of a substance is consumed or produced in a reaction.

In our context of determining partial pressure, the stoichiometric principles allow us to relate the masses of gases to their moles. Once we find the moles, we can use them to determine mole fractions and partial pressures.

Stoichiometry also helps in balancing chemical equations, understanding limiting reactants, and calculating yields. Grasping these concepts is vital for anyone studying chemistry, as they form the foundation for many chemical calculations, making predictions about reactions possible.