assume air sample is 78%N2 and 21%O2 how many units g/mol of the 0.316 g of air
100 g air contains 78 g N2 and 21 g O2.
Hence
Mass of N2 in 0.316 g air = (0.316 g x 78) / 100 = 0.2465 g
Mass of O2 in 0.316 g air = (0.316 g x 21) / 100 = 0.0664 g
Molarmass of N2 =28 g/mol
No. of moles of N2 = 0.2465 g/28 g/mol = 8.80 x 10-3 mol
Molarmass of O2 =32 g/mol
No. of moles of O2 = 0.0664 g/32 g/mol = 2.08 x 10-3 mol
Total no. of moles in 0.316 g air = 8.80 x 10-3 mol + 2.08 x 10-3 mol = 10.88 x 10-3 mol
No. of moles = mass in grams / Molarmass
So molar mass of 0.316 g air = mass in grams / No. of moles = 0.316 g / 10.88 x 10-3 mol = 0.0290 x 103
= 29.0 g/mol
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