
Considering air as a $ 4:1 $ mixture of $ {N_2} $ and $ {O_2} $ , What is the density of air at $ {28^\circ }C $ and 1 atm?
A. $ 1.18{\text{ g}}{{\text{L}}^{ - 1}} $
B. $ 11.62{\text{ g}}{{\text{L}}^{ - 1}} $
C. $ {\text{28}}{\text{.8 g}}{{\text{L}}^{ - 1}} $
D. $ 1.29{\text{ g}}{{\text{L}}^{ - 1}} $
Answer
494.1k+ views
Hint: The ideal gas law is also known as the general gas equation and is the related state variable that describes the state of matter under a given physical condition like pressure, temperature, volume, etc of a hypothetical ideal gas. It is an approximation of behavior under many conditions of many gases. The laws dealing with ideal gases are called ideal gas laws and the laws are intended by the observational work of Boyle and Charles in the seventeenth and eighteenth century respectively.
Complete answer:
The ideal gas equation is PV = nRT
Where n is the ratio of given mass by molecular mass.
$ n = \dfrac{m}{M} $
So ideal gas equation can be written as $ PM = \dfrac{m}{V}RT $
Density is equal to mass per unit volume so $ \dfrac{m}{V} = D $
$ PM = DRT $ ….(I)
According to the question 4 parts of $ {N_2} $ are present in 5 parts of the mixture while 1 part of $ {O_2} $ is present in 5 parts of the mixture.
So, the effective molar mass of $ {N_2} $ $ = \dfrac{4}{5} \times (14 \times 2) \Rightarrow 22.4{\text{ g/mol}} $
(14 is the molecular mass of nitrogen)
Similarly, the effective molar mass of $ {O_2} = \dfrac{1}{5} \times (16 \times 2) \Rightarrow 6.4{\text{ g/mol}} $
Now the effective molar mass of the mixture $ = 22.4 + 6.4 = 28.8{\text{ g/mol}} $
From the given information in the question:
$ T = {28^\circ }C $ in kelvin $ 28 + 273 = 301K $
$ R $ (gas constant) $ = 0.0821{\text{ L atm/mole K}} $
P = 1 atm
$ M = 28.8{\text{ g/mol}} $
Substituting the given values in equation (I)
$ 1 \times 28.8 = D \times 0.0821 \times 301 $
$ D = \dfrac{{28.8}}{{24.71}} $
$ D = 1.166{\text{ }} \approx 1.18{\text{g/litre}} $
Therefore the correct answer is option A.
Note:
One should not get confused with molecular mass or a mass number and molar mass. Molar mass or molecular weight is the sum of the total mass of all the atoms that make up a mole of molecules in grams. The mass carried in one mole of a substance is the molar mass. The unit required to measure molar mass is grams per mole.
Complete answer:
The ideal gas equation is PV = nRT
Where n is the ratio of given mass by molecular mass.
$ n = \dfrac{m}{M} $
So ideal gas equation can be written as $ PM = \dfrac{m}{V}RT $
Density is equal to mass per unit volume so $ \dfrac{m}{V} = D $
$ PM = DRT $ ….(I)
According to the question 4 parts of $ {N_2} $ are present in 5 parts of the mixture while 1 part of $ {O_2} $ is present in 5 parts of the mixture.
So, the effective molar mass of $ {N_2} $ $ = \dfrac{4}{5} \times (14 \times 2) \Rightarrow 22.4{\text{ g/mol}} $
(14 is the molecular mass of nitrogen)
Similarly, the effective molar mass of $ {O_2} = \dfrac{1}{5} \times (16 \times 2) \Rightarrow 6.4{\text{ g/mol}} $
Now the effective molar mass of the mixture $ = 22.4 + 6.4 = 28.8{\text{ g/mol}} $
From the given information in the question:
$ T = {28^\circ }C $ in kelvin $ 28 + 273 = 301K $
$ R $ (gas constant) $ = 0.0821{\text{ L atm/mole K}} $
P = 1 atm
$ M = 28.8{\text{ g/mol}} $
Substituting the given values in equation (I)
$ 1 \times 28.8 = D \times 0.0821 \times 301 $
$ D = \dfrac{{28.8}}{{24.71}} $
$ D = 1.166{\text{ }} \approx 1.18{\text{g/litre}} $
Therefore the correct answer is option A.
Note:
One should not get confused with molecular mass or a mass number and molar mass. Molar mass or molecular weight is the sum of the total mass of all the atoms that make up a mole of molecules in grams. The mass carried in one mole of a substance is the molar mass. The unit required to measure molar mass is grams per mole.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

