Question

The weight of 1 L of ozonised oxygen at STP was found to be 1.5 g. When 100 mL of this mixture at STP was treated with turpentine oil, the volume was reduced to 90 mL. The molecular weight of ozone is :

• A. 49
• B. 47
• C. 49.3
• D. 47.9

Answer 1

The correct option is C 49.3

$1$ L corresponds to $\frac{1}{22.4}=0.0446$ moles at STP. This weighs $1.5$ g, hence, the molar mass of the mixture is $\frac{W}{n}=\frac{1.5}{0.0446}=33.63$ g/mol.
Turpentile oil dissolves ozone. Hence, out of $100$ ml of the mixture, $10$ ml corresponds to ozone and $90$ ml corresponds to oxygen. Since, the volume is directly proportional to the number of moles, out of $0.0446$ moles of the mixture, $0.00446$ moles corresponds to ozone and $0.040$ moles corresponds to oxygen.
Let $M$ be the molar mass of ozone.
The molar mass of the mixture is $\frac{M\times 0.00446+32\times 0.040}{0.0446}=33.63$.
$M=\frac{1.5-1.28}{0.00446}=49.3$ g/mol.