A 20.0 mL solution containing 0.2 g impure $$H_2O_2$$ reacts completely with 0.316 g of $$KMnO_4$$ in acid solution. The purity of $$H_2O_2$$ (in %) is __________ (mol. wt. of $$H_2O_2$$ = 34; mol. wt. of $$KMnO_4$$ = 158)

We are told that a 20.0 mL sample of an impure $$H_2O_2$$ solution contains 0.2 g of the impure reagent and that this entire 0.2 g reacts completely with 0.316 g of $$KMnO_4$$ in acidic medium. Our goal is to find what fraction of that 0.2 g is actually pure $$H_2O_2$$, and then convert that fraction into a percentage.

First, let us calculate the amount (in moles) of $$KMnO_4$$ that took part in the reaction. The molar mass of $$KMnO_4$$ is given as 158.

We have: $$\text{moles of }KMnO_4=\frac{\text{mass}}{\text{molar mass}}=\frac{0.316\;\text{g}}{158\;\text{g mol}^{-1}}$$

Simplifying the fraction: $$\frac{0.316}{158}=0.00200$$

So, $$n(KMnO_4)=0.00200\;\text{mol}$$.

Next, we write the redox reaction between $$KMnO_4$$ and $$H_2O_2$$ in acidic solution:

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

From this balanced equation, the stoichiometric ratio is:

$$2\;\text{mol }KMnO_4 \;\longrightarrow\; 5\;\text{mol }H_2O_2$$

Therefore, $$1\;\text{mol }KMnO_4 \;\longrightarrow\; \frac{5}{2}=2.5\;\text{mol }H_2O_2$$.

Using this ratio, the moles of $$H_2O_2$$ that actually reacted are:

$$n(H_2O_2)=2.5 \times n(KMnO_4)=2.5 \times 0.00200=0.00500\;\text{mol}$$

Now we find the mass corresponding to these moles of pure $$H_2O_2$$. The molar mass of $$H_2O_2$$ is given as 34.

$$\text{mass of pure }H_2O_2 = n \times \text{molar mass} = 0.00500\;\text{mol} \times 34\;\text{g mol}^{-1}=0.170\;\text{g}$$

This 0.170 g of pure $$H_2O_2$$ was present in the original impure sample that weighed 0.200 g. Hence the purity percentage is:

$$\text{Purity }(\%)=\frac{\text{mass of pure }H_2O_2}{\text{mass of impure sample}}\times100 =\frac{0.170}{0.200}\times100=85\%$$

So, the answer is $$85\%$$.