To solve the problem of finding the percentage by mass of \( H_2O_2 \) in a 10 g sample that decolorized 100 ml of 0.1 M \( KMnO_4 \) in acidic medium, we can follow these steps:
### Step 1: Determine the moles of \( KMnO_4 \)
First, we need to calculate the number of moles of \( KMnO_4 \) used in the reaction.
\[
\text{Moles of } KMnO_4 = \text{Molarity} \times \text{Volume (in L)}
\]
Given that the molarity is 0.1 M and the volume is 100 ml (which is 0.1 L):
\[
\text{Moles of } KMnO_4 = 0.1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.01 \, \text{mol}
\]
### Step 2: Determine the equivalent of \( KMnO_4 \)
In acidic medium, \( KMnO_4 \) changes from \( Mn^{7+} \) to \( Mn^{2+} \), which means it gains 5 electrons per mole. Therefore, the number of equivalents of \( KMnO_4 \) can be calculated as:
\[
\text{Equivalents of } KMnO_4 = \text{Moles} \times \text{n-factor}
\]
Here, the n-factor for \( KMnO_4 \) is 5:
\[
\text{Equivalents of } KMnO_4 = 0.01 \, \text{mol} \times 5 = 0.05 \, \text{equivalents}
\]
### Step 3: Determine the equivalents of \( H_2O_2 \)
Since \( H_2O_2 \) also acts as a reducing agent, it will also provide 1 equivalent per mole. Therefore, the equivalents of \( H_2O_2 \) that reacted will be equal to the equivalents of \( KMnO_4 \):
\[
\text{Equivalents of } H_2O_2 = 0.05 \, \text{equivalents}
\]
### Step 4: Calculate the mass of \( H_2O_2 \)
The molecular weight of \( H_2O_2 \) is approximately 34 g/mol. To find the mass of \( H_2O_2 \) that corresponds to 0.05 equivalents:
\[
\text{Moles of } H_2O_2 = \text{Equivalents} = 0.05 \, \text{mol}
\]
\[
\text{Mass of } H_2O_2 = \text{Moles} \times \text{Molar Mass} = 0.05 \, \text{mol} \times 34 \, \text{g/mol} = 1.7 \, \text{g}
\]
### Step 5: Calculate the percentage by mass of \( H_2O_2 \)
Now, we can calculate the percentage by mass of \( H_2O_2 \) in the 10 g sample:
\[
\text{Percentage by mass} = \left( \frac{\text{Mass of } H_2O_2}{\text{Total mass of sample}} \right) \times 100
\]
\[
\text{Percentage by mass} = \left( \frac{1.7 \, \text{g}}{10 \, \text{g}} \right) \times 100 = 17\%
\]
### Final Answer
The percentage by mass of \( H_2O_2 \) in the sample is **17%**.
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