Question

The mass of one litre of a sample of ozonized oxygen at NTP was found to be $1.5$ g. When $100$ mL of this mixture at NTP were treated with turpentine oil, the volume was reduced to $90$ mL, hence, the molar mass (in gm) of ozone is (write the answer as the interger at ten's digit place) :

Answer 1

Volume absorbed by turpentine oil = $10$ mL
$\therefore$ Volume of ozone =$10$ mL (turpentine oil absorbs ozone)
Volume of $O_2=100-10=90$ mL
Now molar mass of ozonized oxygen
=$\frac{wRT}{PV}=\frac{1.5\times 0.0821\times 273}{1\times 1}=33.62g{mol}^{-1}$
Volume ratio of ozone and $O_2$ in 1 litre sample =$100:900$ or mole ratio of ozone and $O_2$ in sample =$100:900$,i.e., $10:90$
$\therefore$ Molar mass of ozonized oxygen=$\frac{10\times molarmassofozone+90\times molarmassofoxygen}{100}$
$33.62=\frac{10\times m_{O_3}+90\times 32}{100}$
Molar mass of $O_3=48.2g{mol}^{-1}$
So, the answer is $4$.