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Sagot :
- Pressure=p=1atm
- Temperature=T=273K.
- V=1dm^3=1L
we need no of moles
[tex]\\ \tt\hookrightarrow PV=nRT[/tex]
[tex]\\ \tt\hookrightarrow n=\dfrac{PV}{RT}[/tex]
[tex]\\ \tt\hookrightarrow n=\dfrac{1}{273(8.314)}[/tex]
[tex]\\ \tt\hookrightarrow n=1/2269.7=0.0004mol[/tex]
No of molecules=No of moles×Avagadro no
[tex]\\ \tt\hookrightarrow 0.0004(6.023\times 10^{23})[/tex]
[tex]\\ \tt\hookrightarrow 2.4\times 10^{20} molecules [/tex]
Hi there!
Note: 1 mole of a any gas at ST and Pressure (273 K and 1 atm) occupies a volume of 22.4 dm^3
=> x mole of oxygen occupies 1.00 dm^3
Therefore, the mole of oxygen on 1.00 dm^3 is :
[tex]mole \: of \: oxygen = \frac{1.00 {dm}^{3} \times 1 \: mole}{22.4 \: dm^{3} } = 0.044642857 \: mole[/tex]
Also note that Avogadros Constant says: 1 mole of any gas contains [tex]6.022 \times {10}^{23}[/tex]molecules.
Hence, 0.044642857 mole of oxygen will contain :
[tex]0.044642857 \times 6.022 \times {10}^{23} = 2.69 \times {10}^{22} \: molecules[/tex]
Therefore, [tex]2.69 \times {10}^{22} \: molecules \: are \: present \: in \: 1dm^{3} \: of \: oxygen \: at \: stp[/tex]
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