Volume strength of $1.5 N H_{2} O_{2}$ is :

4.8

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8.4

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3.0

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8.0

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Solution

The correct option is B
8.4

Volume strength is defined as the term which is used to express the concentration of $H_{2} O_{2}$ in terms of volumes of oxygen gas based on its decomposition to form water and oxygen.
Decomposition reaction:
$H_{2} O_{2} \to 2 H_{2} O + O_{2}$
Let us assume that the solution of M moles of $H_{2} O_{2}$ is present in 1L solution.
2 moles of $H_{2} O_{2}$ gives 1 mol of oxygen
we can also say that 2 moles of $H_{2} O_{2}$ produces 22.4L of oxygen gas at STP.
1 mole $H_{2} O_{2}$ will produce= $\frac{22.4 L}{2} \times 1 = 11.2 L$ of oxygen gas.
Hence, M mole of $H_{2} O_{2}$ will produce (11.2 x M) L of oxygen gas.
So, 1L of $H_{2} O_{2}$ will give (11.2 x M) L of oxygen gas.
therefore, the volume strength of $H_{2} O_{2}$ =11.2 x M
n factor of $H_{2} O_{2}$ is 2
we know that,
N=nM
normality of $H_{2} O_{2}$ =1.5N
So, volume strength of $H_{2} O_{2}$ = $11.2 \times \frac{N}{2} = 11.2 \times \frac{1.5}{2} = 8.4$

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