The volume strength of 1.5 N H_2O_2 solution is:

The correct answer is option 4. 8.4.

The volume strength of \(H_2O_2\) is the term that denotes the concentration of \(H_2O_2\) in terms of the volume of oxygen that is formed alongside water during the decomposition of hydrogen peroxide. The equation for the decomposition of hydrogen peroxide is:

\(2H_2O_2 \rightarrow 2H_2O + O_2\)

We are given the normality of the \(H_2O_2\) solution as 1.5 N. To determine the volume strength, we will first calculate the strength of the 1.5 N \(H_2O_2\) solution. The strength is given by the formula:

\(\text{Strength of H}_2\text{O}_2 = \text{Normality} \times \text{Equivalent weight} (H_2\text{O}_2)\)

The equivalent weight of\(H_2O_2\) is 17 g/mol. Therefore, the strength of the 1.5 N \(H_2O_2\) solution is:

\(\text{Strength of 1.5 N H}_2\text{O}_2 = 1.5 \times 17 = 25.5\)

Next, we will calculate the amount of \(H_2O_2\) using stoichiometry. From the balanced equation, we know that 2 moles of \(H_2O_2\) produce 1 mole of \(O_2\). Since the molar mass of \(H_2O_2\) is 34 g/mol, the mass of 2 moles of \(H_2O_2\) is 68 g/mol.

We also know that 68 g/mol of \(H_2O_2\) produces 22.4 L of oxygen gas at standard temperature and pressure (STP).

Finally, we can calculate the volume strength by dividing the volume of oxygen at STP by the amount of \(H_2O_2\):

\(\text{Volume Strength} = \frac{22.4 \text{ L}}{68 \text{ g/mol}} \times 25.5\)

\(\text{Volume Strength} = 8.4 \text{ L}\)

Therefore, the volume strength of the 1.5 N \(H_2O_2\) solution is 8.4 liters.