Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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) \(3: 16\)

Short Answer

Expert verified
The ratio of molecules is \(7: 32\), option (b).

Step by step solution

01

Understand the Weight Ratio

The problem states that the gases \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) are present in a weight ratio of \(1: 4\). This means that for every 1 unit of weight of \(\mathrm{O}_{2}\), there are 4 units of weight of \(\mathrm{N}_{2}\).
02

Find the Molar Masses

Calculate the molar weights for \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\). The molar mass of \(\mathrm{O}_{2}\) is \(32\, \text{g/mol}\) and the molar mass of \(\mathrm{N}_{2}\) is \(28\, \text{g/mol}\).
03

Set Up the Mole Ratio Equation

Since we are given the weight ratio and asked to find the molecule (or mole) ratio, use the formula: \(\text{Number of moles} = \frac{\text{Weight}}{\text{Molar Mass}}\).
04

Calculate the Number of Moles

Determine the moles of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) using their weight ratio and molar masses:\[n_{\text{O}_2} = \frac{1}{32}\]\[n_{\text{N}_2} = \frac{4}{28}\]
05

Simplify the Mole Ratio

Simplify the mole ratio \(\frac{n_{\text{O}_2}}{n_{\text{N}_2}}\) using the calculated moles:\[\frac{\frac{1}{32}}{\frac{4}{28}} = \frac{1 \times 28}{4 \times 32} = \frac{28}{128} = \frac{7}{32}\]
06

Choose the Correct Option

The simplified ratio of the number of molecules (moles) in the gaseous mixture is \(7: 32\). Therefore, the correct answer is option (b) \(7: 32\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass
Molar mass plays a critical role in stoichiometry as it helps determine the number of moles from a given mass. It’s like a handy conversion tool for chemists. Every substance has a molar mass, which is the mass of one mole of its atoms or molecules. For example, the molar mass of
  • Oxygen molecule \(\mathrm{O}_{2}\) is \(32\, \text{g/mol}\) because oxygen has an atomic mass of \(16\, \text{g/mol}\), and a molecule contains two oxygen atoms.
  • Nitrogen molecule \(\mathrm{N}_{2}\) is \(28\, \text{g/mol}\) given that nitrogen has an atomic mass of \(14\, \text{g/mol}\).
When you know the weight and molar mass of a substance, you can easily find the number of moles. This is crucial when dealing with chemical reactions or mixtures, as it helps predict how substances will interact.
Molecular Ratio
In chemistry, understanding molecular ratios is key to comparing amounts of different substances in a mixture. A molecular ratio reveals how many units of each molecule are present in that mixture. It links the weight ratio (mass ratio) of substances to the number of molecules. To find the molecular ratio, you use the relationship:
  • \(\text{Number of moles} = \frac{\text{Weight}}{\text{Molar Mass}}\)
This allows conversion from a weight-based ratio to a molecule-based ratio. Simplifying this helps in visualizing the relative quantities of each molecule in a mixture, making it easier to predict how they might chemically combine or react. The molecular ratio provides insight into the behavior of compounds in reactions or light mixtures containing gases, solids, or liquids.
Gaseous Mixture
A gaseous mixture is a combination of different gases that coexist without reacting chemically. Think of it as a cocktail of molecules floating around in the air. In the case of a gaseous mixture of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\), these gases remain as separate entities but occupy the same space.When dealing with gaseous mixtures, especially in stoichiometry, it’s important to differentiate between the physical presence of molecules and their potential interactions. Knowing the molecular ratio of gases like \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) can help determine how much of each gas is available, aiding in predictions about chemical behavior or efficiency in chemical processes like combustion.Studying gaseous mixtures enhances the understanding of atmospheric chemistry and processes like respiration, where different gases perform distinct roles but coexist in a mixture.
Number of Moles
The concept of "number of moles" is fundamental in stoichiometry and helps in quantifying chemical substances. Rather than talking about individual atoms or molecules, which are incredibly tiny, chemists use moles to describe amounts of substance in a manageable mathematical way. One mole contains Avogadro’s number, \(6.022 \times 10^{23}\), of entities (atoms, molecules, ions).Given that:
  • \(\text{Number of moles} = \frac{\text{Weight}}{\text{Molar Mass}}\)
This formula allows you to calculate how many moles you have based on the mass of a substance and its molar mass. Calculating the number of moles makes it easier to predict and balance chemical equations. It’s like scaling a recipe—it helps ensure you have the right amount of each ingredient to get the desired "reaction." This concept is especially valuable in predicting yields and understanding proportions in chemical reactions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free