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. $(\text{Take}R=\frac{25}{3}\times {10}^{-2}\text{atmL/molK})$

• A. Molecular weight of the sample is 33.35 g/mol
• B. Molecular weight of ozonized oxygen is 33.35 g/mol
• C. Molecular weight of the sample is 45.5 g/mol
• D. Molecular weight of ozonized oxygen is 45.5 g/mol

Answer 1

Answer: (A), (D)

The correct options are
A Molecular weight of the sample is 33.35 g/mol
D Molecular weight of ozonized oxygen is 45.5 g/mol

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