The volume occupied by \[4.4g\] of \[C{O_2}\] at STP is:
A. \[22.4L\]
B. \[2.24L\]
C. \[0.224L\]
D. \[0.1L\]
Answer
647.4k+ views
Hint: The \[C{O_2}\] stands for carbon dioxide. It is a gas composed of carbon and oxygen. STP stands for standard temperature and pressure.
Complete step by step answer:
The carbon dioxide contains the elements carbon and oxygen in the ration \[1:2\]. The atomic mass of carbon is \[12amu\] and the atomic mass of oxygen is \[32amu\]. The molar mass of \[C{O_2}\] is equal $to = \text{atomic mass of carbon} + 2 \text{(atomic mass of oxygen)}$
\[ = 12 + 32 = 44{\text{ }}g\].
The number of moles contained by \[44g\] of \[C{O_2}\] gas will be equal to \[1mole\]. The given weight of \[C{O_2}\] is \[4.4g\]. The number of moles of contained by \[4.4g\] of \[C{O_2}\] is equal to the ratio of weight of \[C{O_2}\] and molecular mass of \[C{O_2}\].
\[\dfrac{{4.4}}{{44}} = 0.1mole\]
$\implies$ \[1mole\] of \[C{O_2}\] at STP contains \[22.4L\] of the gas. So the \[0.1mole\] of \[C{O_2}\] at STP contains
$\therefore$ \[22.4L \times 0.1 = 2.24L.\]
So, the correct answer is Option B.
Note:
Avogadro’s law states that under the same conditions of temperature and pressure, equal volumes of all gases contain equal numbers of molecules. The equation relating the volume with pressure and temperature is obtained from the kinetic theory of gases of an ideal gas. For real gases the law is applicable at low pressures and high temperatures.
According to Avogadro’s hypothesis, the pressure and temperature at STP are \[101.325{\text{ }}kPa\] and \[273.15{\text{ }}K\] . The equation of volume is equal to
$v = \dfrac{{RT}}{P}$
where \[v\] = volume occupied by one mole of gas
\[R\] = gas constant = \[8.314{\text{ }}J/K/mol\]
\[T\] = temperature
\[P\] = pressure.
Therefore, $v = \dfrac{{8.314 \times 273.15}}{{101.325}}$
$\implies v = 22.4L$
Hence volume for 1 mol of gas and that volume is always equal to the $22.4L$.
Complete step by step answer:
The carbon dioxide contains the elements carbon and oxygen in the ration \[1:2\]. The atomic mass of carbon is \[12amu\] and the atomic mass of oxygen is \[32amu\]. The molar mass of \[C{O_2}\] is equal $to = \text{atomic mass of carbon} + 2 \text{(atomic mass of oxygen)}$
\[ = 12 + 32 = 44{\text{ }}g\].
The number of moles contained by \[44g\] of \[C{O_2}\] gas will be equal to \[1mole\]. The given weight of \[C{O_2}\] is \[4.4g\]. The number of moles of contained by \[4.4g\] of \[C{O_2}\] is equal to the ratio of weight of \[C{O_2}\] and molecular mass of \[C{O_2}\].
\[\dfrac{{4.4}}{{44}} = 0.1mole\]
$\implies$ \[1mole\] of \[C{O_2}\] at STP contains \[22.4L\] of the gas. So the \[0.1mole\] of \[C{O_2}\] at STP contains
$\therefore$ \[22.4L \times 0.1 = 2.24L.\]
So, the correct answer is Option B.
Note:
Avogadro’s law states that under the same conditions of temperature and pressure, equal volumes of all gases contain equal numbers of molecules. The equation relating the volume with pressure and temperature is obtained from the kinetic theory of gases of an ideal gas. For real gases the law is applicable at low pressures and high temperatures.
According to Avogadro’s hypothesis, the pressure and temperature at STP are \[101.325{\text{ }}kPa\] and \[273.15{\text{ }}K\] . The equation of volume is equal to
$v = \dfrac{{RT}}{P}$
where \[v\] = volume occupied by one mole of gas
\[R\] = gas constant = \[8.314{\text{ }}J/K/mol\]
\[T\] = temperature
\[P\] = pressure.
Therefore, $v = \dfrac{{8.314 \times 273.15}}{{101.325}}$
$\implies v = 22.4L$
Hence volume for 1 mol of gas and that volume is always equal to the $22.4L$.
Recently Updated Pages
10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

Trending doubts
Draw a labelled diagram of the neuron and describe class 11 biology CBSE

Explain why Helium is used for filling weather bal class 11 chemistry CBSE

Derive an expression for maximum height and range of class 11 physics CBSE

What do you mean by filtration Explain with a diag class 11 chemistry CBSE

Describe various types of epithelial tissues with the class 11 biology CBSE

Fill in the map of Punjab Before 1947 AD i Chamkaur class 11 social science CBSE

