Question

What will be the strength of $20$ vol of ${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}$in terms of gram per litre?
$60.71\;{\rm{g}}\;{{\rm{L}}^{ - 1}}$
$5.6\;{\rm{g}}\;{{\rm{L}}^{ - 1}}$
$30.62\;{\rm{g}}\;{{\rm{L}}^{ - 1}}$
$17\;{\rm{g}}\;{{\rm{L}}^{ - 1}}$

Answer 1

Volume strength of hydrogen peroxide is the amount of volume of oxygen released on its decomposition. For hydrogen peroxide, ${\rm{20}}\,{\rm{V}}$in ${\rm{1}}\,{\rm{L}}$ means that after the decomposition of ${\rm{1}}\,{\rm{L}}$ peroxide, ${\rm{20}}\,{\rm{V}}$oxygen gas is evolved.

Complete step by step solution:
${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}$ is miscible with water in all proportions and forms a hydrate ${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{.}}{{\rm{H}}_{\rm{2}}}{\rm{O}}$ $\left( {{\rm{mp \,221K}}} \right)$. Hydrogen peroxide has a non-planar structure. In ${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}$ structure in gas phase, dihedral angle is ${\rm{11}}{{\rm{0}}^{\rm{o}}}$ while in solid phase at 110K, the dihedral angle is ${\rm{90}}{\rm{.}}{{\rm{2}}^{\rm{o}}}$.

It acts as an oxidising as well as reducing agent in both acidic and alkaline media.
Hydrogen peroxide is stored in vessels made up of plastic because hydrogen peroxide decomposes in presence of light. To prevent this decomposition of hydrogen peroxide urea is added.
Hydrogen peroxide is used as a disinfectant. It also acts as an antiseptic. It has a wide range of applications in the food industry and pharmaceuticals’ is also used in the manufacture of detergents.
${\rm{2}}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}} \to 2{{\rm{H}}_{\rm{2}}}{\rm{O}} + {{\rm{O}}_{\rm{2}}}$

Hydrogen peroxide contains two hydrogen atoms and two oxygen atoms. So, the molecular mass of hydrogen peroxide
$= 2 \times 1 + 2 \times 16\\\ = 34\;{\rm{gm}}$
Here, we have two moles of hydrogen peroxide, so mass
$= 34 \times 2\\\ = 68\;{\rm{gm}}$
From mole concept we know that ${\rm{22}}{\rm{.4}}\;{\rm{L}}$ of ${{\rm{O}}_{\rm{2}}}$at normal temperature and pressure(NTP) is produced from ${\rm{68}}\;{\rm{gm}}\,{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}$
So, ${\rm{20}}\;{\rm{L}}$of ${{\rm{O}}_{\rm{2}}}$ at NTP is produced from ${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}$;
$= \dfrac{{68 \times 20}}{{22.4}} = 60.7$
Therefore, the strength of ${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}$in 20 volume $${{\rm{H}}{\rm{2}}}{{\rm{O}}{\rm{2}}}$$$= 60.7{\rm{g/l}}$

**Hence, option A is correct.

Note: **
A 30% solution of ${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}$ is marketed as ‘100 volume’ hydrogen peroxide. It means that one millilitre of 30% ${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}$ solution will give ${\rm{100}}\;{\rm{ml}}$ of oxygen at STP. Commercially marketed sample is ${\rm{10}}\;{\rm{V}}$, which means that ${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}$ sample contains 3% ${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}$.