JEE Advanced Numericals on VOLUME STRENGTH OF HYDROGEN PEROXIDE
Problem 1: Complex H₂O₂ Decomposition with Air
Statement:
10 mL of a hydrogen peroxide solution is mixed with 350 mL of air containing 21% O₂ by volume. After complete catalytic decomposition of H₂O₂, the total volume of gases at the same temperature and pressure is 378 mL. Determine the volume strength of the original H₂O₂ solution.
Solution:
Let the volume strength be ( x ) V.
From 10 mL H₂O₂, O₂ produced = ( 10x ) mL.
Initially:
O₂ in air = ( 0.21 \times 350 = 73.5 ) mL
N₂ in air = ( 350 – 73.5 = 276.5 ) mL (inert)
After decomposition:
Additional O₂ from H₂O₂ = ( 10x ) mL
Total O₂ = ( 73.5 + 10x ) mL
N₂ unchanged = 276.5 mL
Total gas volume = ( (73.5 + 10x) + 276.5 = 350 + 10x ) mL
Given:
$$
350 + 10x = 378 \quad\Rightarrow\quad 10x = 28 \quad\Rightarrow\quad x = 2.8
$$
Answer: Volume strength = 2.8 V
Problem 2: Mixing H₂O₂ Solutions with Different Strengths
Statement:
Two hydrogen peroxide solutions of volume strengths 15 V and 5 V are mixed to produce a solution of strength 10 V. Find the ratio of their volumes mixed.
Solution:
Let volumes mixed be ( V_1 ) mL (15 V) and ( V_2 ) mL (5 V).
Moles of H₂O₂ are additive:
[
\frac{15}{11.2} \times V_1 + \frac{5}{11.2} \times V_2 = \frac{10}{11.2} \times (V_1 + V_2)
]
Multiply through by 11.2:
[
15V_1 + 5V_2 = 10(V_1 + V_2)
]
[
15V_1 + 5V_2 = 10V_1 + 10V_2
]
[
5V_1 = 5V_2 \quad\Rightarrow\quad \frac{V_1}{V_2} = \frac{1}{1}
]
Answer: Mixing ratio = 1:1
Problem 3: Oleum Analysis with Percentage Labeling
Statement:
An oleum sample labeled as “112% H₂SO₄” is given. This means 100 g of this oleum when treated with water gives 112 g of pure H₂SO₄. Calculate the percentage of free SO₃ in the oleum.
Solution:
Let 100 g oleum contain ( x ) g free SO₃ and ( (100 – x) ) g H₂SO₄.
Reaction: ( SO_3 + H_2O \rightarrow H_2SO_4 )
Moles of SO₃ = ( \frac{x}{80} )
Mass of H₂SO₄ formed from SO₃ = ( \frac{x}{80} \times 98 = 1.225x ) g
Water consumed = ( \frac{x}{80} \times 18 = 0.225x ) g
Total H₂SO₄ after hydration = ( (100 – x) + 1.225x = 100 + 0.225x ) g
Given:
[
100 + 0.225x = 112 \quad\Rightarrow\quad 0.225x = 12 \quad\Rightarrow\quad x = \frac{12}{0.225} = 53.33
]
Answer: Free SO₃ = 53.33%
Problem 4: Available Chlorine in Bleaching Powder Using Iodometric Titration
Statement:
2.0 g of a bleaching powder sample is treated with excess KI in acidic medium. The liberated iodine requires 40 mL of 0.5 M Na₂S₂O₃ solution. Calculate the percentage of available chlorine in the sample.
Solution:
Reactions:
[
OCl^- + 2I^- + 2H^+ \rightarrow I_2 + Cl^- + H_2O
]
[
I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-}
]
Moles of ( S_2O_3^{2-} ) = ( 0.5 \times 0.04 = 0.02 ) mol
Moles of ( I_2 ) = ( \frac{0.02}{2} = 0.01 ) mol
Moles of ( OCl^- ) = 0.01 mol ≡ 0.01 mol Cl₂ (available chlorine)
Mass of Cl₂ = ( 0.01 \times 71 = 0.71 ) g
[
\% \text{ available chlorine} = \frac{0.71}{2.0} \times 100 = 35.5\%
]
Answer: Available chlorine = 35.5%
