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given
N2 and O2 mixture i.e N2+O2 density = 1.3g/l at stp
i.e 1 lit = 1.3g
therefore 22.4 lit = 22.4 * 1.3
= 29.12 g
we know that at 22.4 lit the weight is known as molecular weight
therfore mw(M)= 29.12 g
wkt.,n=w/M
w= n*M
for given eq., N2 +O2
total weight at 22.4 lit (stp) is = 28x + 32 (1-x)
=> 29.12 = 28x+32-32x
therefore x = 0.72
therefore no.of moles of N2= 0.72 and O2 = 1- 0.72 = 0.28
therefore partial pressure of O2 at stpPO2 = n O2/(n O2 + n N2 ) * p total
p O2 = 0.28/(0.28+0.72) * 1atm
p O2 = 0.28 atm.
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