A 20.0mL solution containing 0.2g impure H2O2 reacts completely with 0.316g of KMnO4 in acid solution. The purity of H2O2(.in %.) is (mol. wt. of H2O2=34' mole wt. of KMnO4=158 )

Question

A 20.0mL solution containing 0.2g impure H2O2 reacts completely with 0.316g of KMnO4 in acid solution. The purity of H2O2(.in %.) is (mol. wt. of H2O2=34' mole wt. of KMnO4=158 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

H2O2+KMnO4 arrow Mn+ 2+O2 [moles of H2O2 ] × 2=(0 . 316/158)× 5 moles of H2O2=5× 10- 3 mass of H2O2=170× 10- 3g % purity =(170 × 10- 3/0 . 2)× 100=85 %

Q. 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 (mol. wt. of $H_{2} O_{2} = 3 4^{^{'}}$ mole wt. of $K M n O_{4} = 158$ )

$H_{2} O_{2} + K M n O_{4} \to M n^{+ 2} + O_{2}$
[moles of $H_{2} O_{2}$ ] $\times 2 = \frac{0.316}{158} \times 5$
moles of $H_{2} O_{2} = 5 \times 1 0^{- 3}$
mass of $H_{2} O_{2} = 170 \times 1 0^{- 3} g$
$%$ purity $= \frac{170 \times 1 0 ^{- 3}}{0.2} \times 100 = 85%$

Q. A 20.0mL solution containing 0.2g impure H2​O2​ reacts completely with 0.316g of KMnO4​ in acid solution. The purity of H2​O2​(in%) is (mol. wt. of H2​O2​=34′ mole wt. of KMnO4​=158 )

H2​O2​+KMnO4​→Mn+2+O2​ [moles of H2​O2​ ] ×2=1580.316​×5 moles of H2​O2​=5×10−3 mass of H2​O2​=170×10−3g % purity =0.2170×10−3​×100=85%

Q. 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 (mol. wt. of $H_{2} O_{2} = 3 4^{^{'}}$ mole wt. of $K M n O_{4} = 158$ )

$H_{2} O_{2} + K M n O_{4} \to M n^{+ 2} + O_{2}$
[moles of $H_{2} O_{2}$ ] $\times 2 = \frac{0.316}{158} \times 5$
moles of $H_{2} O_{2} = 5 \times 1 0^{- 3}$
mass of $H_{2} O_{2} = 170 \times 1 0^{- 3} g$
$%$ purity $= \frac{170 \times 1 0 ^{- 3}}{0.2} \times 100 = 85%$