Question:
Calculate % by mass of a $\text{H}_{2}\text{O}_{2}$ solution that is 67.2 by volume.
Calculate % by mass of a $\text{H}_{2}\text{O}_{2}$ solution that is 67.2 by volume.
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Shortcut Formula: $% \text{ strength} = \frac{17}{56} \times \text{Volume Strength}$.
Here: $\frac{17}{56} \times 67.2 = 17 \times 1.2 = 20.4%$. Faster than finding molarity!
Here: $\frac{17}{56} \times 67.2 = 17 \times 1.2 = 20.4%$. Faster than finding molarity!
Updated On: Jun 19, 2026
- 13.60% by mass
- 20.40% by mass
- 22.44% by mass
- 17.60% by mass
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
We are given the concentration of an aqueous hydrogen peroxide ($\text{H}_2\text{O}_2$) solution in terms of "volume strength" (67.2 V). We need to convert this into a percentage by mass (% w/w).
Step 2: Detailed Explanation:
The "volume strength" of an $\text{H}_2\text{O}_2$ solution refers to the volume of oxygen gas (in liters at STP) produced by the total thermal decomposition of 1 liter of the solution.
1. Find Molarity from Volume Strength:
The standard relationship between Volume Strength ($V$) and Molarity ($M$) for $\text{H}_2\text{O}_2$ is:
$\text{Volume Strength} = 11.2 \times \text{Molarity}$
$67.2 = 11.2 \times M$
$M = \frac{67.2}{11.2} = 6 \text{ M}$ (moles per liter).
This means exactly 1 Liter (1000 mL) of the solution contains 6 moles of $\text{H}_2\text{O}_2$.
2. Calculate the mass of the solute ($\text{H}_2\text{O}_2$):
The molar mass of $\text{H}_2\text{O}_2 = (2 \times 1) + (2 \times 16) = 34 \text{ g/mol}$.
Mass of $\text{H}_2\text{O}_2$ solute = $6 \text{ moles} \times 34 \text{ g/mol} = 204 \text{ g}$.
3. Calculate the % by mass:
To find the percentage by mass (% w/w), we need the total mass of the solution. Standard problems of this type inherently assume the density of the solution is approximately $1 \text{ g/mL}$ (since it is primarily water).
Therefore, the mass of 1 Liter (1000 mL) of the solution is approximately 1000 g.
$\text{% by mass} = \left( \frac{\text{Mass of Solute}}{\text{Mass of Solution}} \right) \times 100%$
$\text{% by mass} = \left( \frac{204 \text{ g}}{1000 \text{ g}} \right) \times 100%$
$\text{% by mass} = 0.204 \times 100% = 20.4%$
Step 3: Final Answer:
The % by mass is 20.40%, matching option (b).
We are given the concentration of an aqueous hydrogen peroxide ($\text{H}_2\text{O}_2$) solution in terms of "volume strength" (67.2 V). We need to convert this into a percentage by mass (% w/w).
Step 2: Detailed Explanation:
The "volume strength" of an $\text{H}_2\text{O}_2$ solution refers to the volume of oxygen gas (in liters at STP) produced by the total thermal decomposition of 1 liter of the solution.
1. Find Molarity from Volume Strength:
The standard relationship between Volume Strength ($V$) and Molarity ($M$) for $\text{H}_2\text{O}_2$ is:
$\text{Volume Strength} = 11.2 \times \text{Molarity}$
$67.2 = 11.2 \times M$
$M = \frac{67.2}{11.2} = 6 \text{ M}$ (moles per liter).
This means exactly 1 Liter (1000 mL) of the solution contains 6 moles of $\text{H}_2\text{O}_2$.
2. Calculate the mass of the solute ($\text{H}_2\text{O}_2$):
The molar mass of $\text{H}_2\text{O}_2 = (2 \times 1) + (2 \times 16) = 34 \text{ g/mol}$.
Mass of $\text{H}_2\text{O}_2$ solute = $6 \text{ moles} \times 34 \text{ g/mol} = 204 \text{ g}$.
3. Calculate the % by mass:
To find the percentage by mass (% w/w), we need the total mass of the solution. Standard problems of this type inherently assume the density of the solution is approximately $1 \text{ g/mL}$ (since it is primarily water).
Therefore, the mass of 1 Liter (1000 mL) of the solution is approximately 1000 g.
$\text{% by mass} = \left( \frac{\text{Mass of Solute}}{\text{Mass of Solution}} \right) \times 100%$
$\text{% by mass} = \left( \frac{204 \text{ g}}{1000 \text{ g}} \right) \times 100%$
$\text{% by mass} = 0.204 \times 100% = 20.4%$
Step 3: Final Answer:
The % by mass is 20.40%, matching option (b).
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