Question

50ml of certain sample of H2O2 gives 1200ml of O2 at STP.The volume strength of H2O2 is

Answer 1

Volume or percentage strength of hydrogen peroxide is a term to express concentration of H2O2 in terms of volumes of oxygen gas based on its decomposition to form water and oxygen.

2H2O2 = 2 H2O + O2

Suppose we have a solution of M moles of H2O2 in 1L solution

From the reaction stoichiometry, 2 moles of H2O2 gives 1mol of oxygen or 22.4L at S.T.P

1mole of H2O2 = 11.2 L of O2

M mole of H2O2 = 11.2 x M L of O2

1L of H2O2 gives 11.2 x M L of O2

Vol strength of H2O2 = 11.2 x M

In terms of Normality, N

n factor for H2O2 is 2

N = M x n or N = M x 2

V S = 11.2 M = 11.2 x N/2 = 5.6 N

Answer 2

Given:

Volume of $\[{H_2}{O_2} = 50\,ml\]$

Volume of $\[{O_2} = 1200\,ml\]$

To Find: Volume strength of $\[{H_2}{O_2}\]$

Solution:

The volume strength of $\[{H_2}{O_2}\]$ solution is the volume of oxygen liberated by the decomposition of $1 ml$ of $\[{H_2}{O_2}\]$  at STP.

If $\[50\,ml\]$ of $\[{H_2}{O_2}\]$ sample gives $\[1200\,ml\]$ of Oxygen at STP, therefore,

$\[1\,ml\]$ of $\[{H_2}{O_2}\]$ solution gives $\[\frac{{1200\,ml}}{{50\,ml}} = 24\,ml\]$ of Oxygen.

Therefore, the volume strength of the sample is $\[24\,V\]$.

Hence, the volume strength of $\[{H_2}{O_2}\]$ is $\[24\,V\]$.

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