Animated Solution for Chemistry - Basic Concepts in Chemistry: The strengths of $5.6$ volume hydrogen peroxide (of density $1\text{ g/mL}$) in terms of mass percentage and molarity (M), respectively, are (Take molar mass of hydrogen peroxide as $34\text{ g/mol}$)

A Deep Dive into Hydrogen Peroxide Concentration

Hydrogen peroxide ($H_2{O}_2$) is a fascinating chemical, often sold in pharmacies in dark bottles. But when you look at the label, instead of seeing a standard concentration like Molarity, you often see a mysterious term: Volume Strength.

Imagine you are holding a beaker containing exactly $1\text{ Liter}$ of an $H_2{O}_2$ solution. If you were to completely decompose this entire liter of liquid into water and oxygen gas ($2H_2{O}_2\to 2H_{2}O+O_2$), the volume of the oxygen gas collected at Standard Temperature and Pressure (STP) is what we call the "Volume Strength". So, a "$5.6\text{ V}$" solution means $1\text{ Liter}$ of it will produce exactly $5.6\text{ Liters}$ of $O_2$ gas.

While volume strength is great for practical laboratory use, in stoichiometry, we prefer Molarity ($M$). Fortunately, there is a beautiful, direct bridge between the two. Because $1\text{ mole}$ of $H_2{O}_2$ produces $0.5\text{ moles}$ of $O_2$ (which occupies $11.2\text{ Liters}$ at STP), we get our master equation:

This is a high-yield concept for competitive exams. Let's substitute our given value into this equation:

We have successfully unlocked the first part of the puzzle: the molarity of the solution is $0.5\text{ M}$.

Now, the problem asks us to shift gears and find the Mass Percentage ($w/w\%$). The formula for mass percentage is fundamentally simple:

To find these masses, we need a canvas to work on. Since mass percentage is an intensive property (it doesn't change whether you have a drop or a bucket of the solution), we can assume any convenient volume. Let's assume we have exactly $1\text{ Liter}$ ($1000\text{ mL}$) of this solution.

First, let's find the mass of this $1000\text{ mL}$ solution. We are given the density ($\rho$) as $1\text{ g/mL}$.

Next, we need the mass of the solute (the actual $H_2{O}_2$ molecules swimming in the water). Since we assumed $1\text{ Liter}$ of solution, and we already know the molarity is $0.5\text{ M}$, it means we have exactly $0.5\text{ moles}$ of $H_2{O}_2$ in our beaker.

To convert moles to mass, we multiply by the molar mass of $H_2{O}_2$, which is given as $34\text{ g/mol}$.

Finally, we bring it all together into our mass percentage formula:

The question specifically asks for the mass percentage and molarity respectively. Therefore, our final sequence is $1.7$ and $0.5$, which perfectly matches option (c).