Answer :
We start by noting that at room temperature and pressure the molar volume of a gas is approximately
[tex]$$
V_m = 24.0 \, \text{dm}^3/\text{mol}.
$$[/tex]
Given a volume of
[tex]$$
V = 6.00 \, \text{dm}^3,
$$[/tex]
we can calculate the number of moles of carbon dioxide ([tex]$\text{CO}_2$[/tex]) gas using the formula
[tex]$$
n = \frac{V}{V_m} = \frac{6.00 \, \text{dm}^3}{24.0 \, \text{dm}^3/\text{mol}} = 0.25 \, \text{mol}.
$$[/tex]
Next, we use Avogadro's number,
[tex]$$
N_A = 6.02214076 \times 10^{23} \, \text{molecules/mol},
$$[/tex]
to convert the number of moles into the number of molecules:
[tex]$$
\text{Number of molecules} = n \times N_A = 0.25 \, \text{mol} \times 6.02214076 \times 10^{23} \, \text{molecules/mol}.
$$[/tex]
This yields
[tex]$$
\text{Number of molecules} \approx 1.50553519 \times 10^{23} \, \text{molecules}.
$$[/tex]
Thus, there are approximately
[tex]$$
1.50553519 \times 10^{23}
$$[/tex]
molecules of carbon dioxide in 6.00 dm³ at room temperature and pressure.
[tex]$$
V_m = 24.0 \, \text{dm}^3/\text{mol}.
$$[/tex]
Given a volume of
[tex]$$
V = 6.00 \, \text{dm}^3,
$$[/tex]
we can calculate the number of moles of carbon dioxide ([tex]$\text{CO}_2$[/tex]) gas using the formula
[tex]$$
n = \frac{V}{V_m} = \frac{6.00 \, \text{dm}^3}{24.0 \, \text{dm}^3/\text{mol}} = 0.25 \, \text{mol}.
$$[/tex]
Next, we use Avogadro's number,
[tex]$$
N_A = 6.02214076 \times 10^{23} \, \text{molecules/mol},
$$[/tex]
to convert the number of moles into the number of molecules:
[tex]$$
\text{Number of molecules} = n \times N_A = 0.25 \, \text{mol} \times 6.02214076 \times 10^{23} \, \text{molecules/mol}.
$$[/tex]
This yields
[tex]$$
\text{Number of molecules} \approx 1.50553519 \times 10^{23} \, \text{molecules}.
$$[/tex]
Thus, there are approximately
[tex]$$
1.50553519 \times 10^{23}
$$[/tex]
molecules of carbon dioxide in 6.00 dm³ at room temperature and pressure.
