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Byju's Answer

Standard XII

Chemistry

Concentration Terms - An Introduction (w/w etc)

20 ml of H2...

Question

$20$ ml of $H_{2} O_{2}$ after acidification with dil $H_{2} S O_{4}$ . required $30$ ml of N/2 $K M n O_{4}$ for complete oxidation. Calculate the % of $H_{2} O_{2}$ in g/lit.

10.75 g/lit

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11.75 g/lit

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12.75 g/ lit

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13.75 g/lit

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Solution

The correct option is B 12.75 g/ lit
$15$ milli equivalents of $K M n O_{4}$ require $15$ milli equivalents of $H_{2} O_{2}$ .

Hence, $15$ milli equivalents are present in $20$ ml of sample.

Thus, strength is $15 \times \frac{17}{20} = 12.75$ .

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Similar questions

Q. Excess of

K I

and dil.

H_{2} S O_{4}

, were mixed in

50

H_{2} O_{2}

. The

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liberated required

20

ml of

0.1

N a_{2} S_{2} O_{3}

. Find out the strength of

H_{2} O_{2}

g / l i t^{- 1}

Q. A

1

g sample of

H_{2} O_{2}

containing

x

percent

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by weight, requires

x

mL of

K M n O_{4}

solution for complete oxidation in acidic medium. Calculate the normality of

K M n O_{4}

solution as nearest integer.

1

g of

H_{2} O_{2}

solution containing

x %

H_{2} O_{2}

by mass requires

x

mL of

K M n O_{4}

for complete oxidation in acid medium. Calculate the normality of

K M n O_{4}

solution. (write the value to the nearest integer)

Q. A solution of

K M n O_{4}

containing

3

g/L is titrated with a solution of

H_{2} O_{2}

containing

2

g/L.

The volume (in ml) of

K M n O_{4}

required to react with

20

H_{2} O_{2}

solution is: (as nearest integer)

3.4

g sample of

H_{2} O_{2}

solution containing

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by weight requires

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solution for complete oxidation under acidic condition. The normality of

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20 ml of H2O2 after acidification with dil H2SO4. required 30 ml of N/2 KMnO4 for complete oxidation. Calculate the % of H2O2 in g/lit.

The correct option is B 12.75 g/ lit15 milli equivalents of KMnO4 require 15 milli equivalents of H2O2. Hence, 15 milli equivalents are present in 20 ml of sample. Thus, strength is 15×1720=12.75.

$20$ ml of $H_{2} O_{2}$ after acidification with dil $H_{2} S O_{4}$ . required $30$ ml of N/2 $K M n O_{4}$ for complete oxidation. Calculate the % of $H_{2} O_{2}$ in g/lit.

The correct option is B 12.75 g/ lit
$15$ milli equivalents of $K M n O_{4}$ require $15$ milli equivalents of $H_{2} O_{2}$ .

Hence, $15$ milli equivalents are present in $20$ ml of sample.

Thus, strength is $15 \times \frac{17}{20} = 12.75$ .