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Standard XII

Chemistry

Equivalent Mass

0.2 g of a sa...

Question

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

25%

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85%

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65%

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95%

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Solution

The correct option is B
85%

No of equivalent of $H_{2} O_{2}$

= $\frac{1}{100}$ 1N $\times$ 17 = 0.17

%Purity = $\frac{0.17}{0.2}$ $\times$ 100 =85%

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Similar questions

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

H_{2} O_{2}

require

10

mL of

1

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 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}

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

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.

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0.2 g of a sample of H2O2 required 10 ml of NKMnO4 in a titration in the presence ofH2SO4 . Purity of H2O2 is

The correct option is B 85% No of equivalent of H2O2 =11001N×17 = 0.17 %Purity =0.170.2×100 =85%

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

The correct option is B
85%

No of equivalent of $H_{2} O_{2}$

= $\frac{1}{100}$ 1N $\times$ 17 = 0.17

%Purity = $\frac{0.17}{0.2}$ $\times$ 100 =85%