Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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

Some Basic Concepts of Chemistry

Solution:

Volume of $O _3$ in $100 \,mL$ of ozonised $O _2$
$=100-90=10\, mL$ (dissolved in turpentine)
Volume of $O _3$ in $1\, L$ of ozonised $O _2$
$=\frac{10 \times 1000}{100} \times 100\, mL$
Volume of $O _2$ in $1 \,L =1000-100=900\, mL$
Weight of $900 \,mL$ of $O _2$ at $STP =\frac{900 \times 32}{22400}=1.286\, g$
Weight of $100 \,mL$ of $O _3$ at $STP =1.5-1.286$
$=0.214 \,g$
Now $100 \,mL$ of $O _3$ at $STP$ weighs $=0.214\, g$
$22400\, mL$ of $O _3$ at STP weighs $=\frac{0.214 \times 22400}{100}$ $=47.94 \,g$
Molecular weight of $O _3=47.94\, g$