A gaseous mixure of O2 and Kr has a density of 1.104g/l at 435 torr and 300 K.?

Question:what is the mole percent O2 in the mixture?

Answers:
PV = nRT, d = m / V, V = m / d

Pm / d = nRT, for a 100 g sample of the gaseous mixture
[(435 / 760)*100] / 1.104 = n*( 0.08206)*(300)
n = 2.11 moles of mixture

O2 + Kr ----> 2.11 moles mixture

32.0g O2
83.8g Kr

[32.0g O2 / 115.8g mixture]*100 = 27.6% O2 composition per mole of mixture

(2.11 moles mixture)*(.276) = .583 moles O2

actually...i'm just gonna stop here because I don't know what i'm doing anymore haha

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