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

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.