[dgolverk]

Hi,
I'm trying to show that 22.4L of gas has 6.02exp23 molecules.
I know that one atomic unit is 1.6605402exp-27kg and the gas I used was methane and there was 0.18g of it in 280mL at STP. How can I calculate the number of molecules in 22.4L using the above information?
Well I solved it (1 molecule /1.6605402 exp -27 kg)(1kg /1000g)(0.18g /280 ml)( 1000ml / l )(22.4 l) = 8.6747 exp 24 molecules. Although there is some error percentage but that's ok. Just one more question: If I want to find how much molecules there are in the 280mL should I just do this, (1 molecule /1.6605402 exp -27 kg)(1kg /1000g)(0.18g /280 ml)?

Thanks,

[ARGOS++]

Dear Dgolverk;

Could it be of help to you if you follow the way (= different Links) in:
"Avogadro – Loschmidt - Boltzmann”

Good Luck!
                    ARGOS⁺⁺

[Hunt]

For CH4 ,atomic mass = 16 amu

Mass of CH4 molecule = 16 x 1.6605402exp-24 g = 26.56 exp-24 g

Number of molecules of methane = 0.18 g / 26.56 exp-24 g.molecule^-1 = 6.77 exp 21 molecules

These molecules occupy a volume of 0.28 L

6.77 exp 21 molecules ----> 0.28 L
N ----- 22.4 L

N is approximately equal to avogadro's number.

your answer needs to be modified as you assumed that the mass of CH4 molecule is 1 u while it is 16 u. your ans : 8.6747 exp 24 molecules divided by 16 = 5.42 exp 23 molecules at STP

For more accurate answer , you can use exact figures.

[dgolverk]

Thank you very much!