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Question

600 mL of a mixture of O3 and O2 weighs 1 g at NTP. Calculate the volume (mL) of ozone in the mixture.

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Solution

We do this question by volume - mole concept
The given is 600ml of mixture of O3 and O2 that is 0.6 L
And the mixture wieghs 1g at NTP
And we have to calculate the volume of O3
1 mole of gas at NTP has 22.4L of volume
Let's take the volume of O3as x L
So the mole of O3 is x/22.4
And if we calculate the mass of 'x' volume M1
The mole will be M1/48. { Molecular mass of O3 is 48 }
So , M148=x22.4 ( Mole of O3 )
M1=4822.4x=2.143x
If we calculate the volume of O2 = ( 0.6 - x ) L
Mole will be = 0.6x22.4
And if we calculate mass of this volume of O2
M2=3222.4(0.6x)=1.43(0.6x) { molecular mass of O2 = 32 }
M1+M2 = 1 ( given in question )
2.143x + 1.43(0.6 - x) = 1
2.143x + 0.857 - 1.43x = 1
0.713x = .143
x = 0.2 L or 200 ml
Volume of ozone will be 200ml
And volume of oxygen is 400ml

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