/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 If air contains \(\mathrm{N}_{2}... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If air contains \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) in volume ratio \(4: 1\) the average vapour density of air is (1) \(14.5\) (2) \(16.5\) (3) \(14.4\) (4) \(29.0\)

Short Answer

Expert verified
14.4

Step by step solution

01

- Understanding Volume Ratio

Air contains \(\text{N}_2\) and \(\text{O}_2\) in a volume ratio of 4:1. This means that for every 4 volumes of \( \text{N}_2 \), there is 1 volume of \( \text{O}_2\).
02

- Molecular Mass Calculation

Calculate the molecular masses: The molecular mass of \( \text{N}_2 \) is 28 (\text{2} \times 14), and the molecular mass of \( \text{O}_2 \) is 32 (\text{2} \times 16).
03

- Weighted Average Molecular Mass

Find the weighted average molecular mass. Using the ratio 4:1, the formula is: \[ \text{Average Molecular Mass} = \frac{(4 \times 28) + (1 \times 32)}{4+1} \] Calculate this to get: \[ \text{Average Molecular Mass} = \frac{112 + 32}{5} = \frac{144}{5} = 28.8 \]
04

- Vapour Density

Vapour density is half of the molecular mass, thus the vapour density of air is: \[ \text{Vapour Density} = \frac{28.8}{2} = 14.4 \]

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 91Ó°ÊÓ!

Key Concepts

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

Volume Ratio
When dealing with the composition of gases like air, understanding volume ratio is crucial. In this exercise, the volume ratio of \(\text{N}_2\) to \(\text{O}_2\) is given as 4:1. This implies that for every 4 volumes of nitrogen gas, there is 1 volume of oxygen gas. Volume ratio helps us understand the proportion of each gas in a mixture. This ratio is crucial for calculating other properties of the mixture, such as its average molecular mass.
Molecular Mass Calculation
Calculating molecular mass is the next important step. The molecular mass of a compound is found by adding together the atomic masses of all the atoms in its molecules. For nitrogen gas \(\text{N}_2\), each nitrogen atom has a mass of 14. Therefore, the molecular mass of \(\text{N}_2\) is 28 (14 + 14). For oxygen gas \(\text{O}_2\), each oxygen atom has a mass of 16. Consequently, the molecular mass of \(\text{O}_2\) is 32 (16 + 16). These molecular masses are necessary for further calculations, such as finding the weighted average molecular mass.
Weighted Average
To find the average molecular mass of the air mixture, we use the concept of the weighted average. This accounts for the different proportions of nitrogen and oxygen in the air. The formula for the weighted average molecular mass, given the 4:1 ratio, is:
\[ \text{Average Molecular Mass} = \frac{(4 \times 28) + (1 \times 32)}{4 + 1} \]
Using this formula:
- Multiply the molecular mass of \(\text{N}_2\) by 4 (since there are 4 volumes of \(\text{N}_2\)): \(4 \times 28 = 112\)
- Multiply the molecular mass of \(\text{O}_2\) by 1 (since there is 1 volume of \(\text{O}_2\)): \(1 \times 32 = 32\)
- Add these results: \(112 + 32 = 144\)
- Divide by the total number of volumes (5): \( \frac{144}{5} = 28.8 \)
Thus, the average molecular mass of the air mixture is 28.8.
Vapour Density
Finally, let's understand vapour density. Vapour density is defined as half of the molecular mass. Knowing the average molecular mass of the air mixture (which we calculated to be 28.8), we can find the vapour density:
\[\text{Vapour Density} = \frac{28.8}{2} = 14.4 \]
Therefore, the vapour density of air is 14.4. This property is useful in various scientific and industrial applications, including calculating the behaviour of air in different conditions.

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

For a real gas, deviations from ideal gas behaviour are maximum at (1) \(10^{\circ} \mathrm{C}\) and \(5 \mathrm{~atm}\) (2) \(10^{\circ} \mathrm{C}\) and \(2 \mathrm{~atm}\) (3) \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) (4) \(100^{\circ} \mathrm{C}\) and \(2 \mathrm{~atm}\)

According to kinetic theory of gases the average translational energy \((\mathrm{K} \mathrm{E})\) is (1) \(\frac{K T}{2}\) per molecule (2) KT per molecule (3) RT per moleculc (4) \(\frac{3}{2} \mathrm{KT}\) per molecule

A small amount of a foul smelling gaseous organic sulphur compound is added to the commercial LPG for leak detection. The property of this gas which helps in leak detection is (1) Sublimation (2) Molecular vibration (3) Molecular association (4) Diffusion

Some moles of oxygen diffused through a small opening in 18 s. Same number of moles of an unknown gas diffuses through the same opening in \(45 \mathrm{~s}\). Molccular weight of the unknown gas is (1) \((32)^{2} \times \frac{18}{45}\) (2) \((32)^{2} \times \frac{18}{(45)^{2}}\) (3) \((32) \times \frac{(45)^{2}}{(18)^{2}}\) (4) \((32)^{2} \times \frac{45}{18}\)

Consider the following statement (I) Joule Thomson experiment is isoenthalpic as well as adiabatic (II) \(\Lambda\) negative value of \(\mu_{J \mathrm{~T}}\) (Joule Thomson coefficient) corresponds to warming of a gas on expansion (III) The temperature at which neither cooling nor heating effect is observed is known as inverson temperature Which of the above statements are correct? (1) I and II (2) I and II (3) II and III (4) I, II and III

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.