Problem 158 - A Complete Resource Book in Chemistry for JEE Main

Question

The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is \(1: 4\). The ratio of number of their molecule is: (a) \(1: 8\) (b) \(3: 16\) (c) \(1: 4\) (d) \(7: 32\)

Step-by-Step Solution

Answer

The ratio of the number of molecules of oxygen to nitrogen is \(7:32\) (option d).

Step 1: Understanding the Given Ratio

We are given the ratio of the masses of oxygen to nitrogen as \(1:4\). This means for every 1 mass unit of oxygen, there are 4 mass units of nitrogen.

Step 2: Identify Molar Masses

The molar mass of oxygen (O extsubscript{2}) is 32 g/mol and the molar mass of nitrogen (N extsubscript{2}) is 28 g/mol. These values are crucial for converting the mass ratio into a number molecule ratio.

Step 3: Calculate Number of Moles

Using the formula \[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar Mass}} \] Calculate the moles of each gas: - Moles of Oxygen, \( n_{O_2} = \frac{1}{32} \). - Moles of Nitrogen, \( n_{N_2} = \frac{4}{28} \).

Step 4: Simplify the Mole Ratio

Compute the ratio of the number of moles:\[ \text{Ratio of moles of } O_2 \text{ to } N_2 = \frac{\frac{1}{32}}{\frac{4}{28}} \]Simplify the expression:\[ \frac{28}{32 \times 4} = \frac{7}{32} \].

Step 5: Conclusion

The ratio of the number of molecules of oxygen to nitrogen is \( \frac{7}{32} \), which corresponds to option \(d\).

Key Concepts

Molar Mass

In chemistry, the molar mass is a fundamental concept that describes the mass of a given substance (chemical element or chemical compound) divided by its amount of substance. The molar mass is usually expressed in grams per mole (g/mol). This means it tells us how much one mole of a given element or compound weighs. For understanding molar mass, knowing the atomic masses from the periodic table is crucial. Every element has a distinct atomic mass number, typically listed below the element symbol. For instance, the molar mass of nitrogen (N) is 14 g/mol since that's its atomic weight. Since nitrogen gas is diatomic (N₂), the molar mass becomes 28 g/mol (14 x 2). Similarly, oxygen (O) has an atomic weight of 16 g/mol, and since oxygen also exists as a diatomic molecule (O₂), its molar mass is 32 g/mol (16 x 2). These molar masses are vital when converting mass ratios into mole ratios, which is an essential step when calculating how different substances will react with each other in a chemical reaction.

Mole Concept

The mole is a basic unit in the field of chemistry that indicates the amount of a substance. It allows chemists to count particles, such as atoms or molecules, by weighing them. One mole is denoted as having exactly 6.022 x 10²³ particles (Avogadro's number). Understanding how to use moles is important when you're dealing with chemical equations and reactions. For instance, if we're given a mass of a substance, we can determine how many moles are present by dividing the mass by the substance's molar mass.- **Formula:** \[ ext{Number of moles} = \frac{\text{Mass of substance}}{\text{Molar Mass}} \]- So, with 1 g of oxygen and 4 g of nitrogen, using their respective molar masses, we find \( n_{O_2} = \frac{1}{32} \) and \( n_{N_2} = \frac{4}{28} \).By learning to convert mass to moles, you can better understand and predict the outcomes of chemical reactions, as well as how different molecules relate to each other in a mixture.

Molecule Ratio

The molecule ratio helps us understand how many molecules of one substance are present in comparison to another in a chemical mixture. This concept is essential when analyzing the composition of gases and solutions. When you have the masses of two substances, using the respective molar masses enables you to convert these masses into moles. This then allows you to find out how many molecules there are of each component. For example, given the masses of oxygen and nitrogen in this exercise, we calculated the moles of each and thus found that the mole ratio is determined by:- Using mass and molar mass, converts it to moles: - Oxygen: \( n_{O_2} = \frac{1}{32} \) - Nitrogen: \( n_{N_2} = \frac{4}{28} \)- Simplifying: \[ \text{Mole Ratio of } O_2:N_2 = \frac{\frac{1}{32}}{\frac{4}{28}} = \frac{7}{32} \]These calculations show how many times one molecule occurs compared to another, thus providing the molecule ratio. Understanding this concept is highly relevant in predicting how gases mix and react with one another.