1. Write the balanced reaction in acidic medium

In acidic solution, $KMnO_4$ oxidizes $H_2O_2$ as:

$2MnO_4^- + 5H_2O_2 + 6H^+ \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O$

So, the mole ratio is:

$2\text{ mol } KMnO_4 : 5\text{ mol } H_2O_2$

2. Calculate moles of $KMnO_4$ used

Given mass of $KMnO_4 = 0.316\,g$

Molar mass of $KMnO_4 = 158$

$n(KMnO_4)=\frac{0.316}{158}=0.002\text{ mol}$

3. Calculate moles of $H_2O_2$ reacting

Using the stoichiometric ratio:

$2\text{ mol } KMnO_4 \rightarrow 5\text{ mol } H_2O_2$

Therefore,

$n(H_2O_2)=0.002\times \frac{5}{2}=0.005\text{ mol}$

4. Calculate mass of pure $H_2O_2$ present

Molar mass of $H_2O_2 = 34$

$m(H_2O_2)=0.005\times 34=0.17\,g$

5. Find purity of the impure sample

Given total mass of impure sample $=0.2\,g$

$\text{Purity }(\%)=\frac{0.17}{0.2}\times 100=85\%$

6. Final answer

$\boxed{85}$

The derived answer matches the stored correct answer.