To find the molarity of the KMnO4 solution used in the titration of As2O3, we will follow these steps:
### Step 1: Calculate the moles of As2O3
To find the moles of As2O3, we use the formula:
\[
\text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}
\]
Given:
- Mass of As2O3 = 0.1097 g
- Molar mass of As2O3 = 198 g/mol (calculated as \(2 \times 75 + 3 \times 16\))
Calculating the moles:
\[
\text{Moles of As2O3} = \frac{0.1097 \, \text{g}}{198 \, \text{g/mol}} \approx 0.000553 \, \text{mol}
\]
### Step 2: Determine the equivalent of As2O3
The n-factor for As2O3 can be determined from the balanced reaction. Each molecule of As2O3 gives 4 electrons (as it goes from As +3 to As +5). Therefore, the number of equivalents of As2O3 is:
\[
\text{Equivalents of As2O3} = \text{Moles of As2O3} \times n = 0.000553 \, \text{mol} \times 4 = 0.002212 \, \text{equivalents}
\]
### Step 3: Relate the equivalents of As2O3 to KMnO4
In the balanced reaction, 1 equivalent of As2O3 reacts with 1 equivalent of KMnO4. Therefore, the equivalents of KMnO4 used in the reaction are also 0.002212 equivalents.
### Step 4: Calculate the normality of KMnO4
Normality (N) is defined as the number of equivalents per liter of solution. The volume of KMnO4 solution used is 26.10 mL, which we convert to liters:
\[
\text{Volume in liters} = \frac{26.10 \, \text{mL}}{1000} = 0.02610 \, \text{L}
\]
Now, we can calculate the normality of the KMnO4 solution:
\[
N = \frac{\text{Equivalents}}{\text{Volume (L)}} = \frac{0.002212 \, \text{equivalents}}{0.02610 \, \text{L}} \approx 0.0846 \, \text{N}
\]
### Step 5: Calculate the molarity of KMnO4
The n-factor for KMnO4 is 5 (since it changes from Mn +7 to Mn +2). We can relate normality and molarity:
\[
N = M \times n \implies M = \frac{N}{n}
\]
Substituting the values:
\[
M = \frac{0.0846 \, \text{N}}{5} \approx 0.01692 \, \text{M}
\]
### Step 6: Final answer
The molarity of the KMnO4 solution is approximately 0.017 M, which rounds to 0.018 M when considering significant figures.
