
What is volume of carbon dioxide, $C{O_2}$ occupying by
i.5 moles and
ii. 0.5 mole of $C{O_2}$ gas measured at STP.
Answer
507.3k+ views
Hint: Temperature and pressure should be kept to a minimum. The standard temperature is 0 degrees Celsius, or 273.15 degrees Kelvin. 1 Atm, 101.3kPa, or 760 mmHg or torr is the standard pressure. The "ordinary" conditions for calculating gas density and volume are known as STP.
Complete answer:
John Dalton (1766–1844) published the first table of normal atomic weight (atomic mass) in 1805, based on a scheme in which the relative atomic mass of hydrogen was described as 1. These proportional atomic masses were calculated using the stoichiometric proportions of chemical reactions and molecules, which greatly helped their acceptance: a chemist did not need to believe in atomic theory (an unproven belief at the time) to use the tables.
In the International System of Units, the mole is the unit of measurement for the volume of matter. It is known as a set of exactly \[6.02214076 \times {10^{23}}\]particles, which may be atoms, molecules, ions, or electrons. The mole is basically a particle count. The particles being counted are usually chemically similar structures that are uniquely distinct. A solution, for example, could have a certain amount of dissolved molecules that are more or less independent of one another. The constituent particles of a solid, on the other hand, are set and bound in a lattice structure, but they can be separated without losing their chemical identity.
${\text{No}}{\text{. of moles = }}\dfrac{{{\text{Volume of C}}{{\text{O}}_2}{\text{ at STP}}}}{{{\text{Molar Volume at STP}}}}$
${\text{Volume of C}}{{\text{O}}_2}{\text{ at STP = No}}{\text{. of moles}} \times {\text{Molar Volume at STP}}$
Molar volume at STP = 22.4 L
Given
Number of moles = 5
${\text{Volume of C}}{{\text{O}}_2}{\text{ at STP = 5 x 22}}{\text{.4 }}$
$ \Rightarrow {\text{Volume of C}}{{\text{O}}_2}{\text{ at STP = 112 l}}$
Given
Number of moles = 0.5
${\text{Volume of C}}{{\text{O}}_2}{\text{ at STP = }}{\text{.5 x 22}}{\text{.4 }}$
$ \Rightarrow {\text{Volume of C}}{{\text{O}}_2}{\text{ at STP = 11}}{\text{.2 l}}$
Note:
Like chemists, chemical engineers use the unit mole widely, but separate unit multiples may be more suited for industrial use. The SI unit for volume, for example, is the cubic metre, which is much greater than the litre, which is generally found in the chemical laboratory. The numerical value of molarity remains the same as the volume of material is expressed in kmol (1000 mol) in industrial-scale processes.
Complete answer:
John Dalton (1766–1844) published the first table of normal atomic weight (atomic mass) in 1805, based on a scheme in which the relative atomic mass of hydrogen was described as 1. These proportional atomic masses were calculated using the stoichiometric proportions of chemical reactions and molecules, which greatly helped their acceptance: a chemist did not need to believe in atomic theory (an unproven belief at the time) to use the tables.
In the International System of Units, the mole is the unit of measurement for the volume of matter. It is known as a set of exactly \[6.02214076 \times {10^{23}}\]particles, which may be atoms, molecules, ions, or electrons. The mole is basically a particle count. The particles being counted are usually chemically similar structures that are uniquely distinct. A solution, for example, could have a certain amount of dissolved molecules that are more or less independent of one another. The constituent particles of a solid, on the other hand, are set and bound in a lattice structure, but they can be separated without losing their chemical identity.
${\text{No}}{\text{. of moles = }}\dfrac{{{\text{Volume of C}}{{\text{O}}_2}{\text{ at STP}}}}{{{\text{Molar Volume at STP}}}}$
${\text{Volume of C}}{{\text{O}}_2}{\text{ at STP = No}}{\text{. of moles}} \times {\text{Molar Volume at STP}}$
Molar volume at STP = 22.4 L
Given
Number of moles = 5
${\text{Volume of C}}{{\text{O}}_2}{\text{ at STP = 5 x 22}}{\text{.4 }}$
$ \Rightarrow {\text{Volume of C}}{{\text{O}}_2}{\text{ at STP = 112 l}}$
Given
Number of moles = 0.5
${\text{Volume of C}}{{\text{O}}_2}{\text{ at STP = }}{\text{.5 x 22}}{\text{.4 }}$
$ \Rightarrow {\text{Volume of C}}{{\text{O}}_2}{\text{ at STP = 11}}{\text{.2 l}}$
Note:
Like chemists, chemical engineers use the unit mole widely, but separate unit multiples may be more suited for industrial use. The SI unit for volume, for example, is the cubic metre, which is much greater than the litre, which is generally found in the chemical laboratory. The numerical value of molarity remains the same as the volume of material is expressed in kmol (1000 mol) in industrial-scale processes.
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