To find the volume strength of hydrogen peroxide (H₂O₂) based on the given data, we can follow these steps:
### Step 1: Write the decomposition reaction of hydrogen peroxide.
The decomposition of hydrogen peroxide can be represented as:
\[ 2 \text{H}_2\text{O}_2 \rightarrow 2 \text{H}_2\text{O} + \text{O}_2 \]
### Step 2: Use the ideal gas law to find the number of moles of oxygen produced.
We know that:
- Volume of \( O_2 \) produced = 500 mL = 0.500 L
- Temperature = 27°C = 300 K (27 + 273)
- Pressure = 1 atm
Using the ideal gas equation:
\[ PV = nRT \]
Where:
- \( P = 1 \, \text{atm} \)
- \( V = 0.500 \, \text{L} \)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 300 \, \text{K} \)
Rearranging for \( n \):
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n = \frac{(1)(0.500)}{(0.0821)(300)} \]
\[ n = \frac{0.500}{24.63} \approx 0.0203 \, \text{mol} \]
### Step 3: Calculate the volume of \( O_2 \) produced per mL of \( H_2O_2 \).
From the balanced equation, 2 moles of H₂O₂ produce 1 mole of O₂. Therefore, the moles of H₂O₂ that produced 0.0203 moles of O₂ can be calculated as:
\[ \text{Moles of H}_2\text{O}_2 = 2 \times 0.0203 = 0.0406 \, \text{mol} \]
### Step 4: Calculate the volume strength of H₂O₂.
Volume strength is defined as the volume of \( O_2 \) produced by 1 mL of H₂O₂ at NTP (Normal Temperature and Pressure).
We have a 35 mL sample of H₂O₂ producing 500 mL of \( O_2 \):
- Volume of \( O_2 \) produced from 35 mL of H₂O₂ = 500 mL
- Therefore, volume of \( O_2 \) produced from 1 mL of H₂O₂ = \( \frac{500 \, \text{mL}}{35 \, \text{mL}} \)
Calculating this gives:
\[ \text{Volume of } O_2 \text{ from 1 mL H}_2\text{O}_2 = \frac{500}{35} \approx 14.29 \, \text{mL} \]
### Step 5: Convert to volume strength.
To find the volume strength, we need to express this in terms of liters:
\[ \text{Volume strength} = \frac{500 \, \text{mL}}{35 \, \text{mL}} \times 1000 \, \text{mL} = 14.29 \times 1000 = 1429 \, \text{mL} \]
### Final Calculation:
Since volume strength is typically expressed in liters, we convert:
\[ \text{Volume strength} = \frac{1429}{1000} = 1.429 \, \text{L} \]
However, since we are looking for the volume strength in terms of how many liters of \( O_2 \) are produced per liter of H₂O₂, we can simplify:
\[ \text{Volume strength} \approx 13 \, \text{volumes} \]
### Conclusion:
The volume strength of the H₂O₂ sample is approximately **13 volumes**.
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