Question

$1120\text{mL}$ of ozonized oxygen at S.T.P. weigh $1.76\text{g}$. Calculate the volume (in mL) of oxygen in the
ozonized oxygen.

Answer 1

Let, the volume of oxygen in $1120\text{mL}$ of ozonized oxygen be $x\text{mL}$ at S.T.P.
Volume of ozone $=(1120-x)\text{mL}$ at S.T.P.
We known that, mass $=$ vol. of mixture $\times$ density
mass = vol. of oxygen × it’s density + vol. of ozone × it’s density

Density of $O_2$ at STP $=\frac{32}{22400}\text{g/mL}$
Density of $O_3$ at STP $=\frac{48}{22400}\text{g/mL}$

$1.76=x\times \frac{32}{22400}+(1120-x)\times \frac{48}{22400}$
$\Rightarrow 1.76\times 22400=32x+53760-48x$
$\Rightarrow 16x=(53760-39424)\text{mL}$
$\Rightarrow x=896\text{mL}$