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Standard XII

Chemistry

Introduction to Le Chatelier's Principle

20 mL of a so...

Question

$20$ mL of a solution containing $0.2$ g of impure sample of $H_{2} O_{2}$ reacts with $0.316$ g of $K M n O_{4}$ in presence of $H_{2} S O_{4}$ . What is the $%$ purity of $H_{2} O_{2}$ ?

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Solution

The redox reaction is as follows:
$2 {M n O_{4}}^{-} + 5 H_{2} O_{2} \to 5 O_{2} + 2 M n^{2 +}$
The number of moles of $K M n O_{4}$ present $= \frac{0.316}{158} = 2 \times 10^{- 3}$ mol
The number of moles of pure $H_{2} O_{2} = \frac{5}{2} \times 2 \times 10^{- 3} = 5 \times 10^{- 3}$
$0.2$ g of sample contains $0.17$ g of pure $H_{2} O_{2}$ .
Therefore, percentage purity of $H_{2} O_{2}$ $= \frac{0.17}{0.2} \times 100 = 85 %$

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20

mL of a solution containing

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$A 20.0 m L$ solution containing $0.2 g$ impure $H_{2} O_{2}$ reacts completely with $0.316 g$ of $K M n O_{4}$ in acid solution. The purity of $H_{2} O_{2}$ (in%) is (Nearest integer) $(mol.wt.of H_{2} O_{2} = 34 mole.wt.of.KM n O_{4} = 158)$

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20 mL of a solution containing 0.2 g of impure sample of H2O2 reacts with 0.316 g of KMnO4 in presence of H2SO4. What is the % purity of H2O2?

The redox reaction is as follows:2MnO4−+5H2O2→5O2+2Mn2+The number of moles of KMnO4 present =0.316158=2×10−3 molThe number of moles of pure H2O2=52×2×10−3=5×10−30.2 g of sample contains 0.17 g of pure H2O2.Therefore, percentage purity of H2O2 =0.170.2×100=85%

$20$ mL of a solution containing $0.2$ g of impure sample of $H_{2} O_{2}$ reacts with $0.316$ g of $K M n O_{4}$ in presence of $H_{2} S O_{4}$ . What is the $%$ purity of $H_{2} O_{2}$ ?