Question:
600 ml of ozonised oxygen at STP was found to weight one gram. What is the volume of O3 in the ozonoised oxygen ?
(1)               200 ml
(2)               150 ml
(3)               100 ml
(4)               50 ml
Answer:
From the ideal gas law we can find the expression for the molar mass:
pV = nRT n = m/M pV = mRT/M
M = mRT/pV = 1∙8.314∙273/(100000∙0.0006) = 37.8 g/mol
M(O3) = 48
M(O2) = 32
The average molar mass of gas can be expressed as the sum of molar masses multiplied by mole fraction. Let x denote the mole fraction of ozone, so the mole fraction oxygen is 1-x. Now we can write the expression for x: 48x + 32(1-x) = 37.8 x = 0.3625
The volume of ozone is the total volume multiplied by the ozone mole fraction:
V(O3) = x∙V = 217.5 ml ≈ 200 ml So, the correct option is (1).
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