1. Identify the redox reaction

In acidic medium, permanganate oxidizes ferrous ion:

$\ce{MnO4^- + 5Fe^{2+} + 8H+ -> Mn^{2+} + 5Fe^{3+} + 4H2O}$

Mohr's salt provides $\ce{Fe^{2+}}$ ions.

2. Calculate moles of Mohr's salt (and hence $\ce{Fe^{2+}}$)

Given:

• Volume of Mohr's salt solution $= 20.0\,\text{mL} = 0.0200\,\text{L}$
• Molarity $= 0.05\,\text{M}$

So, moles of $\ce{Fe^{2+}}$ are:

$n(\ce{Fe^{2+}})=MV=0.05\times 0.0200=0.0010\,\text{mol}$

3. Use stoichiometry to find moles of $\ce{KMnO4}$ required

From the balanced equation:

$1\text{ mol }\ce{MnO4^-} : 5\text{ mol }\ce{Fe^{2+}}$

Hence,

$n(\ce{KMnO4})=\frac{0.0010}{5}=2.0\times 10^{-4}\,\text{mol}$

4. Calculate volume of $0.01\,\text{M } KMnO_4$ used

Given molarity of $\ce{KMnO4}$:

$M=0.01\,\text{mol L}^{-1}$

So,

$V=\frac{n}{M}=\frac{2.0\times 10^{-4}}{0.01}=2.0\times 10^{-2}\,\text{L}=20\,\text{mL}$

Thus, $20\,\text{mL}$ of $\ce{KMnO4}$ is delivered from the burette.

5. Find volume left in the burette

Initial burette reading is zero, and it is a $50\,\text{mL}$ burette. So initially it contains $50\,\text{mL}$ solution.

After delivering $20\,\text{mL}$, volume left is:

$50-20=30\,\text{mL}$

6. Final answer

$\boxed{30\,\text{mL}}$