To solve the problem, we need to determine the volume of `KMnO4` solution left in the burette after reaching the endpoint of the titration with Mohr's salt solution. We will use the concept of millimoles and the stoichiometry of the reaction involved.
### Step-by-Step Solution:
1. **Identify the Given Data:**
- Molarity of `KMnO4` solution (M1) = 0.01 M
- Volume of Mohr's salt solution (V1) = 20.0 mL
- Molarity of Mohr's salt solution (M2) = 0.05 M
- Initial reading of the burette = 0 mL
2. **Calculate the Millimoles of Mohr's Salt:**
\[
\text{Millimoles of Mohr's salt} = M2 \times V1 = 0.05 \, \text{mol/L} \times 20.0 \, \text{mL} = 1.0 \, \text{mmol}
\]
3. **Determine the Stoichiometry of the Reaction:**
The reaction between `Fe^{2+}` (from Mohr's salt) and `KMnO4` can be represented as:
\[
5 \, Fe^{2+} + MnO_4^{-} \rightarrow 5 \, Fe^{3+} + Mn^{2+}
\]
From the balanced equation, 1 mole of `KMnO4` reacts with 5 moles of `Fe^{2+}`. Therefore, the N-factor for `KMnO4` is 5.
4. **Calculate the Millimoles of `KMnO4` Required:**
Using the stoichiometry, we can find the millimoles of `KMnO4` needed to react with 1.0 mmol of `Fe^{2+}`:
\[
\text{Millimoles of } KMnO4 = \frac{1.0 \, \text{mmol}}{5} = 0.2 \, \text{mmol}
\]
5. **Calculate the Volume of `KMnO4` Solution Used:**
Now, we can calculate the volume of `KMnO4` solution required using its molarity:
\[
\text{Volume of } KMnO4 = \frac{\text{Millimoles}}{\text{Molarity}} = \frac{0.2 \, \text{mmol}}{0.01 \, \text{mol/L}} = 20.0 \, \text{mL}
\]
6. **Determine the Volume of `KMnO4` Left in the Burette:**
Since the initial reading was 0 mL and the total volume of `KMnO4` solution in the burette is unknown, we can assume a certain initial volume (let’s denote it as V_initial). After using 20.0 mL of `KMnO4`, the volume left in the burette will be:
\[
\text{Volume left} = V_{\text{initial}} - 20.0 \, \text{mL}
\]
However, since the problem does not specify the initial volume of `KMnO4`, we can assume a standard initial volume of 50 mL (as the burette reading starts from 0 mL). Thus:
\[
\text{Volume left} = 50.0 \, \text{mL} - 20.0 \, \text{mL} = 30.0 \, \text{mL}
\]
### Final Answer:
The volume of `KMnO4` solution left in the burette after the endpoint is **30 mL** (nearest integer).
