The volume strength of 8.9 M H2O2 solution calculated at 273 K and 1 atm is .......... . (R = 0.0821

Q: Why do we use 11.2 instead of 22.4 in the formula?

Because 2 moles of $H_2O_2$ produce only 1 mole of $O_2$. So, 1 mole of $H_2O_2$ produces 0.5 moles of $O_2$. At STP, 0.5 moles of a gas occupies $0.5 \times 22.4 = 11.2\text{ L}$.

Q: What exactly does '8.9 M' mean in this context?

It means there are exactly 8.9 moles of $H_2O_2$ dissolved in 1 liter of the solution.

Q: Is volume strength always calculated at STP?

Usually yes, but if the question specifies a different temperature or pressure (like 298 K), you must calculate the new molar volume using $PV=nRT$ and adjust the multiplier accordingly.

Q: Why did we round 99.68 to 100?

The question explicitly instructed to round off the final answer to the nearest integer. Since the decimal part .68 is greater than .5, we round up to 100.

Q: Can I use the formula $x = 5.6 \times N$ here?

Yes! The normality $N$ of $H_2O_2$ is $2 \times M$ (since its n-factor is 2). So $x = 5.6 \times (2 \times 8.9) = 11.2 \times 8.9 = 99.68$. Both formulas give the exact same result.

Question

The volume strength of $8.9	ext{ M } 	ext{H}_2	ext{O}_2$ solution calculated at $273	ext{ K}$ and $1	ext{ atm}$ is .......... . ($R = 0.0821	ext{ L atm K}^{-1}	ext{mol}^{-1}$) (rounded off to the nearest integer).

Sigma Percentile · Accepted Answer

### Decoding Volume Strength

Imagine a beaker filled with hydrogen peroxide ($H_2O_2$). Over time, it naturally decomposes and releases oxygen gas ($O_2$) bubbles. The term **'volume strength'** is a very practical way chemists express the concentration of this solution. It simply tells us how many liters of oxygen gas we can collect from just one liter of this $H_2O_2$ solution at standard temperature and pressure (STP).

To understand the math behind it, we must first look at the chemical reaction.

$$2H_2O_2(aq) ightarrow 2H_2O(l) + O_2(g)$$

Notice the stoichiometry: **two moles** of aqueous hydrogen peroxide decompose to give two moles of water and exactly **one mole** of oxygen gas. This $2:1$ ratio is the absolute key to our calculation.

### The Master Equation

The question specifies a temperature of $273	ext{ K}$ and a pressure of $1	ext{ atm}$. These are the classic conditions for Standard Temperature and Pressure (STP). At these conditions, one mole of any ideal gas, including our oxygen, occupies exactly $22.4	ext{ L}$. We can verify this using the ideal gas law:

$$V_m = \frac{RT}{P} = \frac{0.0821 	imes 273}{1} = 22.4	ext{ L}$$

Now, let's build our formula. If we have $1	ext{ L}$ of an $M$ molar solution, it contains exactly $M$ moles of hydrogen peroxide. Since two moles of $H_2O_2$ give one mole of oxygen, $M$ moles will give $\frac{M}{2}$ moles of oxygen.

Multiplying these moles by the molar volume ($22.4	ext{ L}$), we get the beautiful relation for volume strength ($x$):

$$x = \frac{M}{2} 	imes 22.4 = 11.2 	imes M$$

### Final Calculation

Now, we just need to plug in the values. We are given that the molarity of the solution is $8.9	ext{ M}$. Let's substitute this into our derived formula:

$$x = 11.2 	imes 8.9$$

Multiplying $11.2$ by $8.9$ gives us $99.68$. This means one liter of this solution will release almost one hundred liters of oxygen gas!

The question explicitly asks us to round off to the nearest integer. Since $99.68$ is closest to $100$, our final volume strength is **$100$**.

Always remember to check the temperature and pressure in such questions. If the temperature was $298	ext{ K}$ (room temperature), the molar volume would be $24.4	ext{ L}$, and our multiplier would change from $11.2$ to $12.2$!

Animated Solution for Chemistry - Basic Concepts in Chemistry: The volume strength of $8.9 M H_{2} O_{2}$ solution calculated at $273 K$ and $1 atm$ is .......... . ( $R = 0.0821 L atm K^{- 1} mol^{- 1}$ ) (rounded off to the nearest integer).

Enter Numerical Value:

Visualized Solution

\text{Concept of Volume Strength}

\text{Decomposition Reaction}

\text{Molar Volume at STP}

\text{Relating Molarity and Volume Strength}

\text{Substitution}

\text{Calculation}

\text{Rounding Off}

\text{What if conditions change?}

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

Decoding Volume Strength

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

\text{Concept of Volume Strength}

\text{Decomposition Reaction}

\text{Molar Volume at STP}

\text{Relating Molarity and Volume Strength}

\text{Substitution}

\text{Calculation}

\text{Rounding Off}

\text{What if conditions change?}

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

Decoding Volume Strength

The Master Equation

Final Calculation

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