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 O31 mole of gas at NTP has 22.4L of volume Let's take the volume of O3as x LSo 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=2.143xIf we calculate the volume of O2 = ( 0.6 - x ) LMole will be = 0.6−x22.4And if we calculate mass of this volume of O2M2=3222.4(0.6−x)=1.43(0.6−x) { 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 = .143x = 0.2 L or 200 mlVolume of ozone will be 200ml And volume of oxygen is 400ml

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