At STP, the molar volume of an ideal gas is 22.4 L. In various other words, the volume of 1 mol of a gas prefer oxygen is 22.4 L. (mol/22.4L) or (22.4L/mol)

The difficulty gives friend 113.97 l of oxygen. Utilizing dimensional analysis, friend can discover the variety of moles that oxygen.

You are watching: How many oxygen molecules are present in 113.97 liters of oxygen gas at stp?

113.97 l x (1 mol/22.4L) = 5.0879 mol of oxygen.

The problem asks because that the variety of oxygen molecules and also the variety of molecules in a mol is 6.02 x 10^23.

5.0879 mol oxygen x ((6.02 x 10^23) oxygen molecules/ mol oxygen ) = 3.06 x 10^24 molecules


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First let"s figure how many moles that oxygen room present.

At STP, 1 mole rectal 22.4 liters. So,

113.97 liters * (1 mole / 22.4 liters) = 5.09 mole.

Now there are 6.02e23 molecules per mole. So multiply this through the number of moles, us get

(6.02 x 1023 molecule / mole)(5.09 mole) = 3.06 x 1024 molecules.

Hope this helps.


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