Question

$600mL$ of a mixture of $O_3$ and $O_2$ weighs $1g$ at NTP. Calculate the volume (mL) of ozone in the mixture.

Answer 1

We do this question by volume - mole concept

The given is 600ml of mixture of $O_3$ and $O_2$ that is 0.6 L

And the mixture wieghs 1g at NTP

And we have to calculate the volume of $O_3$

1 mole of gas at NTP has 22.4L of volume

Let's take the volume of $O_3$as 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 , $\frac{M_1}{48}=\frac{x}{22.4}$ ( Mole of O3 )

$M_1=\frac{48}{22.4}x=2.143x$

If we calculate the volume of O2 = ( 0.6 - x ) L

Mole will be = $\frac{0.6-x}{22.4}$

And if we calculate mass of this volume of $O_2$

$M2=\frac{32}{22.4}(0.6-x)=1.43(0.6-x)$ { molecular mass of O2 = 32 }

$M_1+M_2$ = 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