Question

Calculate the volume strength of $1.5\mathrm{N}{\mathrm{H}}_2{\mathrm{O}}_2$.

• A. 4.8L
• B. 8.4L
• C. 3.0L
• D. 8.0L

Answer 1

The correct option is B

8.4L

Explanation for correct answer:-

Option (B) 8.4L

Step 1: Introducing the formula

$\mathrm{Strength}=\mathrm{Normality}\mathrm{x}\mathrm{equivalent}\mathrm{weight}$

Step 2: Finding out the equivalent weight

$\mathrm{Equivalent}\mathrm{weight}=\frac{\mathrm{Molar}\mathrm{mass}}{\mathrm{n}-\mathrm{factor}}=\frac{34}{2}=17$

Step 3: Calculating the strength of $1.5\mathrm{N}{\mathrm{H}}_2{\mathrm{O}}_2$

Given that Normality of Hydrogen Peroxide: $1.5\mathrm{N}$

$\mathrm{Strength}\mathrm{of}1.5\mathrm{N}{\mathrm{H}}_2{\mathrm{O}}_2=1.5\times 17=25.5\mathrm{g}/\mathrm{L}$

Step: 4 Calculating the volume strength of $1.5\mathrm{N}{\mathrm{H}}_2{\mathrm{O}}_2$

$2{\mathrm{H}}_2{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{O}}_2$

2 moles of ${\mathrm{H}}_2{\mathrm{O}}_2$ produce 1 mole of ${\mathrm{O}}_2$ and 1 mole of ${\mathrm{H}}_2\mathrm{O}$.

The $\mathrm{mass}\mathrm{of}2\mathrm{moles}\mathrm{of}{\mathrm{H}}_2{\mathrm{O}}_2=68\mathrm{g}/\mathrm{mol}$
Also, 68 g/mol ${\mathrm{H}}_2{\mathrm{O}}_2$ produces 22.4 L of oxygen gas at STP.

Now,
$\mathrm{Volume}\mathrm{strength}=\frac{\mathrm{Volume}\mathrm{of}{\mathrm{O}}_2}{\mathrm{Mass}\mathrm{of}{\mathrm{H}}_2{\mathrm{O}}_2}\times \mathrm{Strength}\mathrm{of}{\mathrm{H}}_2{\mathrm{O}}_2=\left(\frac{22.4}{68}\right)\times 25.5L=8.40\mathrm{L}$
Therefore, the volume strength of the given solution is 8.40L.