Question

The volume of perhydrol which on decomposition gives 1.5 lit of \[{{O}_{2}}\] gas at STP is:
A. 25 ml
B. 15 ml
C. 10 ml
D. 20 ml

Answer 1

Hint: Perhydrol is another name of the chemical compound hydrogen peroxide, with formula ${{H}_{2}}{{O}_{2}}$. Hydrogen peroxide is a liquid that decomposes to produce water and oxygen. Volume strength of perhydrol gives the volume of the oxygen gas that decomposes.

Complete answer:
We have been given perhydrol, which on decomposition produces 1.5 lit of oxygen gas, \[{{O}_{2}}\] at STP. We have to find the volume of hydrogen peroxide needed to yield this amount of volume of oxygen gas. The reaction of decomposition of perhydrol is,
\[{{H}_{2}}{{O}_{2}}\to {{H}_{2}}O+\dfrac{1}{2}{{O}_{2}}\]
We know that at STP condition, Volume strength of 1 L of the perhydrol contains 100 ml of oxygen gas, which means 1 L of perhydrol contains 100 ml of oxygen gas that on decomposition gives oxygen and water. The volume strength of perhydrol and volume of oxygen has a relation,
Volume of \[{{O}_{2}}\]= volume strength $\times $ volume of ${{H}_{2}}{{O}_{2}}$
so, 1.5 lit = 1500 mL of oxygen is yielded by,
1500 = 100 $\times $ volume of ${{H}_{2}}{{O}_{2}}$
Volume of ${{H}_{2}}{{O}_{2}}$= $\dfrac{1500\,mL}{100}$
So, volume of perhydrol = 15 ml
Hence, 15 ml of perhydrol is required to give 1.5 liter of oxygen gas.

Note:
The volume strength of hydrogen peroxide can also be expressed in terms of its moles as, 1 M of perhydrol = 11.2 L of oxygen. This is derived from the STP volume of oxygen as 22.7 L, and from the number of moles of oxygen produced by perhydrol