Problem 153 - A Complete Resource Book in Chemistry for JEE Main

Question

Equal masses of methane and oxygen are mixed in an empty container at \(25^{\circ} \mathrm{C}\). The fraction of the total pressure exerted by oxygen is [2007] (a) \(1 / 3 \times 273 / 298\) (b) \(1 / 3\) (c) \(1 / 2\) (d) \(2 / 3\)

Step-by-Step Solution

Answer

The fraction of the total pressure exerted by oxygen is \(\frac{1}{3}\).

Step 1: Understanding Mass and Molar Mass

First, let's determine the molar mass of both methane (CH₄) and oxygen (O₂). The molar mass of methane is 16 g/mol (C = 12, H = 4). The molar mass of oxygen is 32 g/mol (O = 16 x 2). Since the masses are equal, we have an equal weight of methane and oxygen.

Step 2: Calculate Moles of Each Gas

Since equal masses of gases are used, let's consider 16 g of each gas. For methane: \[ \text{Moles of } \text{CH}_4 = \frac{16 \text{ g}}{16 \text{ g/mol}} = 1 \text{ mol} \] For oxygen: \[ \text{Moles of } \text{O}_2 = \frac{16 \text{ g}}{32 \text{ g/mol}} = 0.5 \text{ mol} \] So, we have 1 mol of methane and 0.5 mol of oxygen.

Step 3: Calculate Total Pressure and Mole Fraction

Total moles of gas in the mixture = 1 mol (CH₄) + 0.5 mol (O₂) = 1.5 mol. The mole fraction of oxygen, which represents the fraction of total pressure it exerts, is given by \[ \text{Mole fraction of } \text{O}_2 = \frac{\text{Moles of } O_2}{\text{Total moles}} = \frac{0.5}{1.5} = \frac{1}{3} \]

Step 4: Choose the Correct Answer

Based on our calculations, the fraction of the total pressure exerted by oxygen is given by option (b) \(\frac{1}{3}\). Therefore, the correct answer is option (b).

Key Concepts

Mole Fraction

The concept of mole fraction is a crucial component when dealing with gas mixtures in chemistry. To put it plainly, the mole fraction is a way of expressing the concentration of a component in a mixture. It reflects how much of the whole mixture a single component comprises.

When calculating the mole fraction, you simply take the number of moles of that particular component and divide it by the total number of moles present in the mixture.

Here's a refresher from our example:

• Total moles of gas = 1 mole of CH₄ + 0.5 mole of O₂ = 1.5 moles
• Mole fraction of O₂ = \( \frac{\text{Moles of } O_2}{\text{Total moles}} = \frac{0.5}{1.5} = \frac{1}{3} \)

The mole fraction doesn’t have any units because it is simply a ratio. Understanding this allows you to predict how each gas in a mixture will contribute to the total pressure, according to Dalton’s Law of Partial Pressures.

Molar Mass Calculation

Understanding molar mass is key to solving problems in chemical mixtures. Molar mass is the weight of one mole of a substance, often expressed in grams per mole (g/mol). Knowing the molar mass allows you to convert between the mass of a substance and the amount in moles.

Here's how we approached the molar mass calculation in our problem:

• The molar mass of methane (CH₄) is determined by adding together the atomic masses of carbon (12) and hydrogen (1 × 4), resulting in 16 g/mol.
• The molar mass of oxygen (O₂) involves adding the atomic masses of two oxygen atoms (16 × 2), which gives 32 g/mol.

Armed with molar mass information, the conversion becomes straightforward. For instance, knowing that the mass and molar mass are equal enables us to calculate the number of moles of each gas swiftly, as displayed in the original exercise.

Gas Laws

Gas laws in chemistry provide a theoretical framework to understand the behavior of gases under various conditions. One of the central principles is how gas components contribute to total pressure in a mixture, as per Dalton’s Law of Partial Pressures.

In essence, each gas in a mixture exerts a pressure (its partial pressure) proportional to its mole fraction in the whole mixture.

• Dalton’s Law: Total pressure exerted by a gas mixture is the sum of the partial pressures of individual gases.
• Partial Pressure: For a specific component, it is the product of its mole fraction and the total pressure of the mixture.

By understanding gas laws, you can predict and calculate how each component in a mixture will behave and contribute to observable properties like pressure and temperature. These laws are not just abstract principles; they are tools chemists use to decode real-world processes involving gases.