How many molecules of $C{O_2}$ are present in one litre of air conditioning \[0.03\% \] by volume of $C{O_2}$ at STP?
A.$8.066 \times {10^{10}}$
B.$8.066 \times {10^{ - 10}}$
C.$8.066 \times {10^{ - 18}}$
D.$8.066 \times {10^{18}}$
Answer
582.9k+ views
Hint:Air is a mixture of gases as it contains nitrogen, oxygen, argon and carbon dioxide. It refers to the earth’s atmosphere. There is also a very minute amount of water vapor in the air.
Formula used:
${\text{Moles = }}\dfrac{{{\text{Volume}}}}{{22.4{\text{ litre}}}}{\text{ = }}\dfrac{{{\text{Number of molecules}}}}{{{\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{23}}}}$
Complete step by step answer:
The given volume of air is one litre. Carbon dioxide is 0.03 percent of the volume of the air. Hence the volume of carbon dioxide present in the given volume of air is as follows,
${\text{Volume of C}}{{\text{O}}_2} = {\text{ Percentage of C}}{{\text{O}}_2}{\text{ }} \times {\text{ Volume of air = }}\dfrac{{0.03}}{{100}} \times 1 = 0.0003{\text{ litres}}$
Hence, the volume of $C{O_2}$ is \[0.0003\] litre.
The relation between the moles of carbon dioxide and its volume is as follows,
${\text{Moles = }}\dfrac{{{\text{Volume of C}}{{\text{O}}_2}}}{{22.4{\text{ litre}}}}{\text{ = }}\dfrac{{0.0003}}{{22.4}}{\text{ = 0}}{\text{.00001339}}$
Molecules of carbon dioxide can be calculated through its moles and that can be done as follows,
${\text{Moles = }}\dfrac{{{\text{Number of molecules of C}}{{\text{O}}_2}}}{{{\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{23}}}}$
${\text{Number of molecules of C}}{{\text{O}}_2} = {\text{ 0}}{\text{.00001339}} \times {\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{23}}$
$ \Rightarrow {\text{Number of molecules of C}}{{\text{O}}_2} = 0.0000806 \times {10^{23}} = 8.06 \times {10^{18}}$
The total number of carbon dioxide molecules is ${\text{8}}{\text{.06}} \times {\text{1}}{{\text{0}}^{18}}$.
Therefore, the correct option is D.
Note:
Carbon dioxide is a colourless gas. Its density is higher than that of dry air. Atmospheric carbon dioxide is the main carbon source for life on the earth. Carbon dioxide is produced by all aerobic organisms. This is released when they metabolize organic compounds to produce energy by respiration. Carbon dioxide in the atmosphere is at low concentration. It is a greenhouse gas. Carbon dioxide in the solid form is known as dry ice. Carbon dioxide can also be found in the deposits of petroleum and natural gas. It possesses a sharp and acidic odour but at normal encountered concentrations, it is odourless. It is also produced during the fermentation of sugars in bread, beer, wine, etc.
Formula used:
${\text{Moles = }}\dfrac{{{\text{Volume}}}}{{22.4{\text{ litre}}}}{\text{ = }}\dfrac{{{\text{Number of molecules}}}}{{{\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{23}}}}$
Complete step by step answer:
The given volume of air is one litre. Carbon dioxide is 0.03 percent of the volume of the air. Hence the volume of carbon dioxide present in the given volume of air is as follows,
${\text{Volume of C}}{{\text{O}}_2} = {\text{ Percentage of C}}{{\text{O}}_2}{\text{ }} \times {\text{ Volume of air = }}\dfrac{{0.03}}{{100}} \times 1 = 0.0003{\text{ litres}}$
Hence, the volume of $C{O_2}$ is \[0.0003\] litre.
The relation between the moles of carbon dioxide and its volume is as follows,
${\text{Moles = }}\dfrac{{{\text{Volume of C}}{{\text{O}}_2}}}{{22.4{\text{ litre}}}}{\text{ = }}\dfrac{{0.0003}}{{22.4}}{\text{ = 0}}{\text{.00001339}}$
Molecules of carbon dioxide can be calculated through its moles and that can be done as follows,
${\text{Moles = }}\dfrac{{{\text{Number of molecules of C}}{{\text{O}}_2}}}{{{\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{23}}}}$
${\text{Number of molecules of C}}{{\text{O}}_2} = {\text{ 0}}{\text{.00001339}} \times {\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{23}}$
$ \Rightarrow {\text{Number of molecules of C}}{{\text{O}}_2} = 0.0000806 \times {10^{23}} = 8.06 \times {10^{18}}$
The total number of carbon dioxide molecules is ${\text{8}}{\text{.06}} \times {\text{1}}{{\text{0}}^{18}}$.
Therefore, the correct option is D.
Note:
Carbon dioxide is a colourless gas. Its density is higher than that of dry air. Atmospheric carbon dioxide is the main carbon source for life on the earth. Carbon dioxide is produced by all aerobic organisms. This is released when they metabolize organic compounds to produce energy by respiration. Carbon dioxide in the atmosphere is at low concentration. It is a greenhouse gas. Carbon dioxide in the solid form is known as dry ice. Carbon dioxide can also be found in the deposits of petroleum and natural gas. It possesses a sharp and acidic odour but at normal encountered concentrations, it is odourless. It is also produced during the fermentation of sugars in bread, beer, wine, etc.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: 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

State and prove Bernoullis theorem class 11 physics CBSE

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

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

