Question

How many molecules of oxygen are there in 1.00 dm3 of oxygen at standard pressure and a temperature of 273 K. Give your answer to three significant digits.​

Answer 1

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 :

$mole \: of \: oxygen = \frac{1.00 {dm}^{3} \times 1 \: mole}{22.4 \: dm^{3} } = 0.044642857 \: mole$

Also note that Avogadros Constant says: 1 mole of any gas contains $6.022 \times {10}^{23}$molecules.

Hence, 0.044642857 mole of oxygen will contain :

$0.044642857 \times 6.022 \times {10}^{23} = 2.69 \times {10}^{22} \: molecules$

Therefore, $2.69 \times {10}^{22} \: molecules \: are \: present \: in \: 1dm^{3} \: of \: oxygen \: at \: stp$

Answer 2

• Pressure=p=1atm
• Temperature=T=273K.
• V=1dm^3=1L

we need no of moles

$\\ \tt\hookrightarrow PV=nRT$

$\\ \tt\hookrightarrow n=\dfrac{PV}{RT}$

$\\ \tt\hookrightarrow n=\dfrac{1}{273(8.314)}$

$\\ \tt\hookrightarrow n=1/2269.7=0.0004mol$

No of molecules=No of moles×Avagadro no

$\\ \tt\hookrightarrow 0.0004(6.023\times 10^{23})$

$\\ \tt\hookrightarrow 2.4\times 10^{20} molecules$