How many molecules are contained in 134.5 L of #O_2# at STP?

1 Answer
Trevor Ryan.
Jan 21, 2016

#3.6138xx10^24#

Explanation:

At STP, 1 mole of all gases occupies a volume of #22.4dm^3#.

Therefore, since #1L=1000cm^3#, we may use ratio and proportion to show that #134.5L=134500cm^3=0.1345m^3=134.5dm^3#.

Therefore this volume contains #134.5/22.4=6mol#.

But 1 mole contains an Avagadro number of particles, ie #6.023xx10^23#.

Therefore by ration and proportion, 6 moles of oxygen will contain #6xx6.023xx10^23=3.6138xx10^24# oxygen molecules.

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