Question

The density of O2 and N2 is 1.4 g/L at stp.Find out the partial pressure of O2 in the mixture

Answer 1

Explanation:

N2 and O2 mixture i.e N2+O2

density = 1.4g/l at stp

i.e 1 lit = 1.4g

therefore 22.4 lit = 22.4 * 1.4 = 31.36 g

We know that at 22.4 lit the weight is known as molecular weight (M)

therfore M= 31.36 g

wkt.,n=w/M

w= n*M

Let weight of N2 be x

Thus for given eq., N2 +O2

Total weight at 22.4 lit (stp) is = 28x + 32 (1-x)

=> 31.36 = 28x+32-32x

therefore x = 0.16

therefore no.of moles of N2= 0.16 and O2 = 1- 0.16 = 0.84

therefore partial pressure of O2 at stp PO2 = n O2/(n O2 + n N2 ) * p total

p O2 = 0.84/(0.16 + 0.84)* 1atm

p O2 = 0.84 atm

Answer 2

Answer:

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