Question

What is the percentage strength (weight/volume) of \[11.2V\]\[{H_2}{O_2}\]?
A. \[1.7\]
B. \[3.4\]
C. \[34\]
D. None of these

Answer 1

Hint: We need to know that the percentage strength is always indicated in a term with a decimal. And here we acquire the number of constituents present in a solution or number of solutes present in a solution or liquid preparation. The amount of concentration is shown in mg/mL. In the case of a solution, the percentage strength indicates the amount of substance dissolved in a particular amount of liquid.

Complete answer:
The percentage strength of \[11.2V\] \[{H_2}{O_2}\] is not equal to \[1.7\]. Hence, option (A) is incorrect.
Hence the option (B) is correct.
Here the percentage strength indicates the amount of oxygen formed by one liter of \[{H_2}{O_2}\]. Hence, we have to produce oxygen from \[{H_2}{O_2}\]. Let’s see the equation,
\[2{H_2}{O_2} \to 2{H_2}O + {O_2}\]
Here, two moles of hydrogen peroxide make one mole of oxygen with two mole of water.
At STP, \[68g\] of \[{H_2}{O_2}\] (molar mass of two moles of \[{H_2}{O_2}\]) make \[22.4L\] of oxygen.
Let us assume, the volume strength is equal to ‘x’
Therefore, x L of oxygen formed from \[\dfrac{{68x}}{{22.4}}gm\] of hydrogen peroxide.
Hence, the volume strength of x L of \[{O_2}\] is \[\dfrac{{68x}}{{22.4}} = \dfrac{{17x}}{{5.6}}\]
Next we have to find out the molarity using molar mass and volume of the solution. Thus, the molar mass of hydrogen peroxide is equal to\[34\] and the volume of the solution is considered as one liter.
Substitute the values in the equation of molarity.
Molarity, M\[ = \dfrac{{moles}}{{volume(L)}}\]
\[\dfrac{{moles}}{{volume(L)}} = \dfrac{{\dfrac{{mass}}{{molarmass}}}}{{volume}}\]
Where, mass is equal to \[\dfrac{{68x}}{{22.4}}gm\]. Hence,
\[\dfrac{{moles}}{{volume(L)}} = \dfrac{{\dfrac{{\dfrac{{68x}}{{22.4}}gm}}{{34}}}}{{1L}} = \dfrac{x}{{11.2}}\]
The given volume strength is equal to \[11.2\]. Therefore,
\[molarity = \dfrac{{11.2}}{{11.2}} = 1\]
Hence, the strength is equal to \[34g/l\]
So, the weight ratio can be find out by using the equation,
\[\% w/w = \dfrac{{massofsolute}}{{massofsolution}} \times 100\]
Now we can substitute the known values we get,
\[ = \dfrac{{34}}{{1000}} \times 100\]
On simplification we get,
$ = 3.4\% $
Thus, the percentage strength is equal to \[3.4\% \]. Hence, option (B) is correct.
The percentage strength of hydrogen peroxide is not equal to \[34\]. Hence, option (C) is incorrect.
The percentage strength (weight/volume) of \[11.2V\] \[{H_2}{O_2}\] is equal to \[3.4\% \]. Hence, option (D) is incorrect.

So, the correct answer is “Option B”.

Note:
As we know , hydrogen peroxide is a chemical compound having the formula \[{H_2}{O_2}\]. And it decomposes very easily. When the hydrogen peroxide is decomposed, there is a formation of oxygen as well as hydrogen gas. But if it decomposes in the vacuum, there will not be a formation of hydrogen gas and only the oxygen and water is formed.