Question

The weight of one litre sample of ozonized oxygen at 1 atm and 276 K was found to be 1.45 g. When 100 mL of this mixture was treated with turpentine oil the volume was reduced to 90 mL. Hence, calculate the molecular weight of ozone. $(\text{Take}R=\frac{25}{3}\times {10}^{-2}\text{atmL/molK})$

Answer 1

Since terpentine oil absorbs $O_3$, the volume of $O_3$ absorbed by terpentine oil is 10 mL.
Volume of $O_2=100-10=90$ mL
From ideal gas equation,
$PV=\frac{WRT}{M}$
where, M = Molecular weight of the ozonized oxygen sample
$M=\frac{WRT}{PV}$
$M=\frac{1.45\times \frac{25}{3}\times {10}^{-2}\times 276}{1\times 1}$
$M=33.35$g/mol
Molar ratio of $O_2$ and $O_3$ = 90:10
$\text{Molecular weight of ozonized oxygen}=\frac{90\times 32+10\times a}{100}=33.35$
$a=45.5$g/mol