Question

Calculate the number of carbon and oxygen atom in 11.2 L of CO2 at NTP.

Answer 1

Given ;
volume = 11.2 L

No. of atoms

= (22.4/11.2) × 6.022 × 10^23

= 3.011 × 10^23 atoms
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Answer 2

Answer: $3.01\times 10^{23}$  atoms of carbon and $6.023\times 10^{23}$  atoms of oxygen.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP and contains avogadro's number $6.023\times 10^{23}$ of particles.

Normal condition of temperature (NTP) is 273 K and atmospheric pressure is 1 atm respectively.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given volume}}{\text{Avogadro's volume}}=\farc{11.2L}{22.4L}=0.5moles$

1 mole of $CO_2$ contains =$1\times 6.023\times 10^{23}=6.023\times 10^{23}$ atoms of carbon.

0.5 moles of $CO_2$ contains =$\frac{6.023\times 10^{23}}{1}\times 0.5=3.01\times 10^{23}$  atoms of carbon.

1 mole of $CO_2$ contains =$2\times 6.023\times 10^{23}=12.04\times 10^{23}$ atoms of oxygen.

0.5 moles of $CO_2$ contains =$\frac{12.4\times 10^{23}}{1}\times 0.5=6.023\times 10^{23}$  atoms of oxygen.