20 ml of H-2O-2 after acidification with dil H-2SO-4- required 30 ml of N-2 KMnO-4 complete oxidation- Calculate the - of H-2O-2 in g-lit-10-75 g-lit12-75 g- lit13-75 g-lit11-75 g-lit

Question

Toppr Admin · Accepted Answer

15 milli equivalents of KMnO4 require 15 milli equivalents of H2O2-xA0-Hence- 15 milli equivalents are present in 20 ml of sample-xA0-Thus- strength is 15-xD7-1720-12-75-

$20$ 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.
10.75 g/lit
12.75 g/ lit
13.75 g/lit
11.75 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 $15 \times \frac{17}{20} = 12.75$ .

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.10.75 g/lit12.75 g/ lit13.75 g/lit11.75 g/lit

15 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.
10.75 g/lit
12.75 g/ lit
13.75 g/lit
11.75 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 $15 \times \frac{17}{20} = 12.75$ .