#### A 20.0 mL solution containing 0.2 g impure H2O2 reacts completely with 0.316 g of KMnO4 in acidic solution. The purity of H2O2 (in%) is _________ (mol.wt.of H2O2 =34, mol. wt. of KMnO4 = 158)Option: 1 85Option: 2 -Option: 3 -Option: 4 -

The reaction will be-

$\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{KmnO}_{4} \rightarrow \mathrm{Mn}^{+2}+\mathrm{O}_{2}$

In above-balanced reaction

Electron transfer of H2O2    (n) = 2  (-2 to 0 of Oxygen)

Electron transfer of KmnO4  (n) = 5  (+7 to +2 of Mn)

We know this formula:

$\text {Number of Eq. of } \mathrm{H}_{2} \mathrm{O}_{2}=\text {Number of Eq. of } \mathrm{KMnO}_{4}$

( Mole X n-factor )  of H2O= ( Mole X n-factor ) of KmnO4

$\mathrm{x} \times 2=\frac{0.316}{158} \times 5$

$\mathrm{x}=5 \times 10^{-3} \mathrm{~mol}$

Now

Mass =  Mole X Molar mass

$\mathrm{mass}_{\mathrm{H}_{2} \mathrm{O}_{2}}=5 \times 10^{-3} \times 34=0.17 \mathrm{gm}$

We know,

$%\text { purity }=\frac{\text { mass of pure product }}{\text { mass of impure product obtained }} \times 100 \%$

$%\text { purity of }\mathrm{H_2O_2} =\frac{0.17}{0.2}\times 100=85 \ %$

Ans = 85