To solve the problem step by step, we will follow these calculations:
### Step 1: Calculate the moles of NaOH in each solution before mixing.
1. **For the first solution (20 mL of 0.2 M NaOH):**
\[
\text{Moles of NaOH} = \text{Molarity} \times \text{Volume (L)} = 0.2 \, \text{mol/L} \times 0.020 \, \text{L} = 0.004 \, \text{mol}
\]
2. **For the second solution (35 mL of 0.1 M NaOH):**
\[
\text{Moles of NaOH} = 0.1 \, \text{mol/L} \times 0.035 \, \text{L} = 0.0035 \, \text{mol}
\]
### Step 2: Calculate the total moles of NaOH after mixing.
\[
\text{Total moles of NaOH} = 0.004 \, \text{mol} + 0.0035 \, \text{mol} = 0.0075 \, \text{mol}
\]
### Step 3: Calculate the concentration of NaOH in the final 100 mL solution.
\[
\text{Concentration (M)} = \frac{\text{Total moles}}{\text{Total volume (L)}} = \frac{0.0075 \, \text{mol}}{0.1 \, \text{L}} = 0.075 \, \text{M}
\]
### Step 4: Calculate the moles of NaOH in 40 mL of the diluted solution.
\[
\text{Moles of NaOH in 40 mL} = 0.075 \, \text{mol/L} \times 0.040 \, \text{L} = 0.003 \, \text{mol}
\]
### Step 5: Determine the moles of oxalic acid reacted.
Oxalic acid (H₂C₂O₄) reacts with NaOH in a 1:2 ratio. Therefore, the moles of oxalic acid that reacted can be calculated as:
\[
\text{Moles of } H_2C_2O_4 = \frac{\text{Moles of NaOH}}{2} = \frac{0.003 \, \text{mol}}{2} = 0.0015 \, \text{mol}
\]
### Step 6: Calculate the mass of pure oxalic acid.
The molar mass of oxalic acid (H₂C₂O₄) is approximately 90 g/mol.
\[
\text{Mass of pure } H_2C_2O_4 = \text{Moles} \times \text{Molar mass} = 0.0015 \, \text{mol} \times 90 \, \text{g/mol} = 0.135 \, \text{g}
\]
### Step 7: Calculate the mass of the impure sample.
Given that the sample is 10% pure, we can set up the equation:
\[
\text{Pure mass} = 0.1 \times \text{Total mass of sample}
\]
Let \( m \) be the total mass of the impure sample.
\[
0.135 \, \text{g} = 0.1 \times m \implies m = \frac{0.135 \, \text{g}}{0.1} = 1.35 \, \text{g}
\]
### Final Answer:
The mass of the impure sample is **1.35 g**.
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