Question

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Answer 1

Answer:

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Explanation:

In 0.2 g of a sample of H2O2,

Let ''x'' gm of pure H2O2 is present, then

Equivalents of H2O2 = Equivalents of KMnO4

moles of H202 X V.F of H202 = moles of KMnO4 X V.F of KMnO4

'''' = Molarity x volume x V.f of KMnO4

'''' = Normality x V.F of KMnO4 [N =M . V.f]

thus,

(x/34) x 2 = 1 x 10/1000

x/17 = 1/100

x = 17/100

x = 0.17.

Thus Pure H202 in 0.2 gm sample is =

0.17/0.2 x 100

= 85 %