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0.2 g of a sample of H2O2 is required for 10 ml of 1 N KMnO4 in a titration in the presence of H2SO4. Purity of H2O2
0.2 g of a sample of H2O2 is
is (a) 25% (b) 85% (c) 65% (d) 95%
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 %
I hpe this correct!
Whenever titration prob. comes try to use equivalents of each ractants and products if it is!
Approve!
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