Question

The strength of $11.2$ volume solution of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ is: [Given that molar mass of ${\text{H = 1gmo}}{{\text{l}}^{{\text{ - 1}}}}$ and ${\text{O = 16gmo}}{{\text{l}}^{{\text{ - 1}}}}$ ]
A.$13.6\% $
B.$3.4\% $
C.$34\% $
D.$1.7\% $

Answer 1

Hint: The amount of solute or the concentration of solute present in the total solution is known as the strength. The term Volume strength indicates the volume of oxygen produced by one liter of hydrogen peroxide.

Complete step by step answer:
Volume strength is nothing but the number of volumes of oxygen liberated by one liter of hydrogen peroxide. This statement seems somewhat typical so we will try to understand it by its derivation.
We have to make oxygen from hydrogen peroxide.
The equation involved is:
${\text{2}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}} \to {\text{2}}{{\text{H}}_{\text{2}}}{\text{O + }}{{\text{O}}_{\text{2}}}$
2 moles of hydrogen peroxide produces one mole of oxygen gas.
So we can say that $68gm$(mass of two moles of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$) of hydrogen peroxide produces $22.4L$ of oxygen at ${\text{STP}}$ .
Let us assume that volume strength is ${\text{x}}$ .
So, $x$ liter of oxygen is produced from $\dfrac{{68x}}{{22.4}}gm$ of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$.
Therefore the volume strength for $x$ liter of oxygen is $\dfrac{{68x}}{{22.4}} = \dfrac{{17x}}{{5.6}}$
We will calculate the molarity now. To calculate molarity we need mass, molar mass and volume of the solution, hence we have calculated the mass i.e. $\dfrac{{17x}}{{5.6}}$ , the molar mass of hydrogen peroxide is $34$ and the total volume of the solution is taken as $1L$ .
Now put all these values in the formula of molarity,
Molarity, ${\text{M = }}\dfrac{{{\text{moles}}}}{{{\text{volume(L)}}}}$
Therefore the Molarity=$\dfrac{{{\text{moles}}}}{{{\text{volume}}}} = \dfrac{{\dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}}}{{{\text{volume}}}} = \dfrac{{\dfrac{{\dfrac{{68x}}{{22.4}}gm}}{{34}}}}{{1L}} = \dfrac{x}{{11.2}}$
We have given that volume strength is $11.2$ .
$ \Rightarrow {\text{Molarity = }}\dfrac{{11.2}}{{11.2}} = 1$
Hence we can say that the strength is $34g/l$ .
Therefore the weight ratio is:
$
  \% w/w = \dfrac{{{\text{mass of solute}}}}{{{\text{mass of solution}}}} \times 100 = \dfrac{{34}}{{1000}} \times 100 \\
  \% w/w = 3.4\% \\
$
Therefore the strength in percentage is $3.4\% $ .
Hence option (B) is correct.

Note:
Hydrogen peroxide decomposes easily. It decomposes itself into two components: water and the oxygen. If it is decomposed in an environment it will form some other compounds made up of hydrogen and oxygen but in vacuum only the water and oxygen molecules are formed.