$50$ mL of pure and dry $O_{2}$ was subjected to silent electric discharge and on cooling to the original temperature the volume of ozonised oxygen was found to be $47$ mL. The gas was then brought in contact with turpentine oil. After the absorption of ozone, the remaining gas occupied a volume of $41$ mL, the molecular mass (in g) of ozone is (all measurements are made at constant P and T) :

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Solution

Let the formula of ozone be $O_{n}$ . Suppose $a$ mL of $O_{2}$ reacts to give ozone.
$n O_{2} \to 2 O_{n}$
Volume before reaction, $50$ mL $0$
Volume after reaction, $(50 - a)$ mL $\frac{2 a}{n}$ mL
$∵$ Volume of $O_{2}$ left = $41$ mL
$∴$ $50 - a$ $=$ $41$ mL or $a$ $=$ $9$
Also, volume of $O_{3}$ formed = $47 - 41$ mL $=$ $6$ mL
$∴ \frac{2 a}{n} = 6 ∴ n = 3$ .
The molecular formula of ozone is $O_{3}$ .
And molar mass of ozone is $48$ g/mol.

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