# What would be the strength of \[20\] volume of ${H_2}{O_2}$?

A) $13.6\;g{L^{ - 1}}$

B) $60.7\;g{L^{ - 1}}$

C) $160\;g{L^{ - 1}}$

D) $20.2\;g{L^{ - 1}}$

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**Hint:**We know that hydrogen peroxide decomposes to form water molecules and liberates oxygen gas and we also know that strength is expressed in terms of weight to volume percentage ratio in a solution or in terms of volume it is given as volume to volume ratio.

**Complete step by step answer:**

As we know that two moles of hydrogen peroxide $({H_2}{O_2})$ when decomposes, results in the formation of two moles of water and one mole of oxygen which can be shown with the help of the equation as:

$2{H_2}{O_2} \to 2{H_2}O + {O_2}$

We can also say that $68g$ of hydrogen peroxide reacts with $32g$ of oxygen or we can say that $22.4L$of oxygen is produced from $68g$ of hydrogen peroxide at the standard conditions of temperature and pressure.

Now, in the question we are given with \[20\] volume of hydrogen peroxide and the term is called volume strength which is basically unique to hydrogen peroxide only and it means that $20L$of oxygen is produced at STP by the decomposition of $1L$ of hydrogen peroxide solution.

So we can calculate the amount of oxygen produced in grams using the formula given below as:

$\dfrac{{mass}}{{molecular\;mass}} = \dfrac{{volume\;in\;L\;at\;STP}}{{22.4L}}$

$ \Rightarrow mass = \dfrac{{20 \times 68}}{{22.4L}}$

$ \Rightarrow mass = 60.7g$

Therefore, the strength of hydrogen peroxide in \[20\] volume of ${H_2}{O_2}$ will be given as the ratio of weight of solute to volume of solution and it will be equivalent to $60.7\;g{L^{ - 1}}$.

**Thus, from the above calculation the correct answer is (B).**

**Note:**Always remember that $x$ volumes of hydrogen peroxide means $x$ litres of oxygen is obtained by decomposition of $1$ litres of hydrogen peroxide. Also, the volume strength of hydrogen peroxide is unique to this compound and it is given as volume of oxygen at STP given by one volume of sample of the hydrogen peroxide.