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

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

we need no of moles

`\\ \tt\hookrightarrow PV=nRT`

`\\ \tt\hookrightarrow n=(PV)/(RT)`

`\\ \tt\hookrightarrow n=(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* 10^(23))`

`\\ \tt\hookrightarrow 2.4* 10^(20) molecules`

Answer 2

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) * 1 \: mole}{22.4 \: dm^(3) } = 0.044642857 \: mole`

Also note that Avogadros Constant says: 1 mole of any gas contains
`6.022 * {10}^(23)`molecules.

Hence, 0.044642857 mole of oxygen will contain :

`0.044642857 * 6.022 * {10}^(23) = 2.69 * {10}^(22) \: molecules`

Therefore,
`2.69 * {10}^(22) \: molecules \: are \: present \: in \: 1dm^(3) \: of \: oxygen \: at \: stp`