The purity of H 2 O 2 in a given sample is 85 - - Calculate the weight of impure sample of H 2 O 2 which requires 10 ml of M -5 KMnO 4 solution in a titration in acidic mediumA- 2 gmsB- 0-2 gmsC- 17-0 gmsD- 1-5 gms

The correct option is B 0.2 gms m_eq of MnO^ _4 = meq of H_2O_2 10 ×M/5× = m_eq of H_2O_2 ⇒ 10 m_eq of H_2O_2 Weight of H_2O_2 = 10 × 10^ 3×34/2× 0.17 =0.2 gms ...

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Standard XII

Chemistry

Hydrogen Peroxide

The purity of...

Question

The purity of $H_{2} O_{2}$ in a given sample is 85%.Calculate the weight of impure sample of $H_{2} O_{2}$ which requires 10 ml of $\frac{M}{5} K M n O_{4}$ solution in a titration in acidic medium

2 gms

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0.2 gms

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17.0 gms

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1.5 gms

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Solution

The correct option is B 0.2 gms
$m_{e q} o f M n O_{4}^{-} = m e q o f H_{2} O_{2}$
$10 \times \frac{M}{5} \times = m_{e q} o f H_{2} O_{2}$
$\Rightarrow 10 m_{e q} o f H_{2} O_{2}$
$W e i g h t o f H_{2} O_{2} = 10 \times 10^{- 3} \times \frac{34}{2} \times 0.17 = 0.2 g m s$
$W e i g h t o f i m p u r e H_{2} O_{2} = \frac{100}{85} \times 0.17 = 0.2 g m s$

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Similar questions

Q. The purity of

H_{2} O_{2}

in a given sample is

85 %

. Calculate the weight of impure sample of

H_{2} O_{2}

which requires

10

mL of

\frac{M}{5} K M n O_{4}

solution in a titration in acidic medium.

0.2 g of a sample of $H_{2} O_{2}$ required 10 ml of N $K M n O_{4}$ in a titration in the presence of $H_{2} S O_{4}$ . Purity of $H_{2} O_{2}$ is

20

mL of a solution containing

0.2

g of an impure sample of

H_{2} O_{2}

reacts with

0.316

g of

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(acidic medium). Calculate % purity of

H_{2} O_{2}

0.2 g of a sample of $H_{2} O_{2}$ required 10 ml of N $K M n O_{4}$ in a titration in the presence of $H_{2} S O_{4}$ . Purity of $H_{2} O_{2}$ is

0.2

g of a sample of

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require

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The purity of H2O2 in a given sample is 85%.Calculate the weight of impure sample of H2O2 which requires 10 ml of M5 KMnO4 solution in a titration in acidic medium

The correct option is B 0.2 gms meq of MnO−4=meq of H2O2 10×M5×=meq of H2O2 ⇒ 10 meq of H2O2 Weight of H2O2=10×10−3×342×0.17=0.2 gms Weight of impure H2O2=10085×0.17=0.2 gms

The purity of $H_{2} O_{2}$ in a given sample is 85%.Calculate the weight of impure sample of $H_{2} O_{2}$ which requires 10 ml of $\frac{M}{5} K M n O_{4}$ solution in a titration in acidic medium