Problem 69 - A Complete Resource Book in Chemistry for JEE Main

Question

A gaseous mixture contains \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) in the ratio \(1: 4\) by weight. Then the ratio of their number of molecules in the mixture is (a) \(3: 32\) (b) \(7: 32\) (c) \(1: 4\) (d) \(3: 16\)

Step-by-Step Solution

Answer

The ratio of their number of molecules is \(7: 32\).

Step 1: Identify Known Ratios

The problem states that the weight ratio of oxygen (\(\mathrm{O}_{2}\)) to nitrogen (\(\mathrm{N}_{2}\)) is \(1:4\). This means for every 1 part by weight of \(\mathrm{O}_{2}\), there are 4 parts by weight of \(\mathrm{N}_{2}\).

Step 2: Determine Molecular Weights

Calculate the molecular weights of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\). The molecular weight of \(\mathrm{O}_{2}\) is \(2 \times 16 = 32 \) g/mol, and for \(\mathrm{N}_{2}\), it is \(2 \times 14 = 28 \) g/mol.

Step 3: Calculate Moles

For \(\mathrm{O}_{2}\): Using the 1 part of weight, \(\frac{1}{32}\) moles of \(\mathrm{O}_{2}\). For \(\mathrm{N}_{2}\): Using 4 parts of weight, \(\frac{4}{28}\) moles of \(\mathrm{N}_{2}\).

Step 4: Simplify Moles

Calculate the actual number of moles for each gas in terms of their ratio: For \(\mathrm{O}_{2}\), \(\frac{1}{32}\) moles. For \(\mathrm{N}_{2}\), \(\frac{4}{28} \approx \frac{1}{7}\) moles.

Step 5: Determine Mole Ratios

To find the number of molecules, use the ratio of the moles: Multiply both by the least common multiple (224) to simplify the fractions: For \(\mathrm{O}_{2}\): \(\frac{1 imes 7}{32 imes 7} = \frac{7}{224}\) For \(\mathrm{N}_{2}\): \(\frac{32}{224}\) moles. Thus, the ratio is \(7: 32\).

Key Concepts

Molecular Weight Calculation

When discussing the molecular weight calculation of gases like oxygen (O_{2}) and nitrogen (N_{2}), the molecular weight is an important factor in converting between grams and moles. Each element has a specific atomic weight, found on the periodic table. To calculate the molecular weight of a compound, such as oxygen which is O_{2}, you simply multiply the atomic weight of one oxygen atom, which is 16 g/mol, by 2 because there are two oxygen atoms per molecule. Hence, the molecular weight of O_{2} is 32 g/mol. Similarly, you do the same for nitrogen. Each N_{2} molecule is made of two nitrogen atoms, with each weighing 14 g/mol, leading to a nitrogen molecular weight of 28 g/mol. Calculating molecular weight helps in determining how many moles you have when you are given a specific weight of the compound. The formula used here: \[ \text{Molecular weight} = \text{Atomic weight} \times \text{Number of atoms} \] allows chemical calculations like the one in this exercise.

Mole Ratio

The mole ratio is a central concept in chemistry that helps relate amounts of reactants and products in chemical reactions. It tells you the ratio in which substances react and are produced. In the exercise, we start with a weight ratio of O_{2} to N_{2} of \(1:4\). Once you have the weights, you convert these to mole quantities using the molecular weights. For our gases, we use \( \frac{1 \, \text{part}}{32 \, \text{g/mol}} \) for O_{2} and \( \frac{4 \, \text{parts}}{28 \, \text{g/mol}}\) for N_{2}. This gives us a mole ratio of approximately \( \frac{1}{32} \text{ moles of } O_{2}\) and \( \frac{1}{7} \text{ moles of } N_{2}\). Understanding mole ratios is crucial when you're trying to find the number of molecules in a mixture, allowing you to predict how substances will react or the amount you'll get of each product and reactant.

Stoichiometry

Stoichiometry is what ties all these concepts together. It allows us to use known quantities and ratios to calculate unknown quantities in chemical reactions. In this exercise, the stoichiometric calculations help determine the molecule ratio in the mixture. Starting from the weight ratio and molecular weights, we converted to moles and derived their ratios. The mole ratio of \( \frac{1}{32} \) for O_{2} and \( \frac{1}{7} \) for N_{2} was expressed as \( 7:32 \) after simplifying the fractions. Using Stoichiometry involves balancing equations and understanding that matter cannot be created or destroyed, only transformed. This principle ensures that calculations such as converting between mass, moles, and molecules will always work out correctly with the constraints of chemistry laws. This understanding is what leads to the confident prediction that for every 7 molecules of O_{2}, there are 32 molecules of N_{2} in this mixture.