Molar volume at STP is 22.4 L/mol. A hydrogen peroxide solution is labelled '11.2 volume', meaning 1 L of it releases 11.2 L of O2 at STP on complete decomposition (2H2O2 -> 2H2O + O2). The volume of 0.1 M acidified KMnO4 needed to oxidise 10 mL of this solution is:

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Molar volume at STP is $22.4 L/mol$ . A hydrogen peroxide solution is labelled '11.2 volume', meaning $1 L$ of it releases $11.2 L$ of $O X_{2}$ at STP on complete decomposition ( $2 H X_{2} O X_{2} 2 H X_{2} O + O X_{2}$ ). The volume of $0.1 M$ acidified $KMnO X_{4}$ needed to oxidise $10 mL$ of this solution is:

11.2 L

O X_{2}

= 0.5 mol, which comes from 1 mol

H X_{2} O X_{2}

, so the solution is

1 M

. In the titration

H X_{2} O X_{2}

has

n = 2

10 \times 1 \times 2 = 20

meq.

KMnO X_{4}

has

N = 0.5

, so

V = 20/0.5 = 40 mL