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Air is approximately 80% nitrogen and 20% oxygen (on a mole basis). If 6 g of hydrogen is added to a 22.4 liter maintained at 0 degrees c and initially filled with air at 1 atm pressure, what will be the molecular mass (i.e. the average molecular mass) of the hydrogen-air mixture to 3 sig figs?

I tried to use PV=nRT to get the number of moles

(1 atm*22.4 L)/(.08206*273 K)= .9998 mol

I multiplied it by 62.04 g/mol (mass of H2 + O2 + N2) and got 62.027 g, but I don't think this is right because I didn't incorporate the composition of air (80% nitrogen and 20% oxygen) into the problem. Could someone help me out?

Ok, here goes:

Change the percents to a mass basis>

.80*molmassN2 + .20*molmassO2= mass air inside.

Figure the mass of the air inside.
Now add the mass of H2.

Now you have the original air, plus H2.
You have .8 mole N2, .2moleO2, and 6/molmassH2. That is about 4 moles of the mixture total.

Take the total mass inside
.8 *molmassN2 +.2molmassO2 + 6/molmassH2

divide by the total "moles" inside (about 4).
That is the "average" mole mass, whatever that means. This reminds me of taking a class of 10 boys and 10 girls, and using statistics to find the average gender.
But, average "mol" mass has some uses.

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