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
654.6k+ 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
Which will be the least stable resonating structure class 11 chemistry CBSE

How many 5 digit telephone numbers can be construc-class-11-maths-CBSE

How do you find the angle of the resultant vector class 11 physics CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

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

What is the need and importance of classification class 11 biology CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

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

10 examples of friction in our daily life

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

Difference between physical and chemical change class 11 chemistry CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

