{"id":162834,"date":"2026-01-22T14:58:00","date_gmt":"2026-01-22T14:58:00","guid":{"rendered":"https:\/\/news.gyankatta.org\/?p=162834"},"modified":"2026-01-22T14:59:09","modified_gmt":"2026-01-22T14:59:09","slug":"jee-advanced-numericals-on-volume-strength-of-hydrogen-peroxide","status":"publish","type":"post","link":"https:\/\/news.gyankatta.org\/?p=162834","title":{"rendered":"JEE Advanced Numericals on VOLUME STRENGTH OF HYDROGEN PEROXIDE"},"content":{"rendered":"\n

<\/p>\n\n\n\n

\n\n\n\n

Problem 1: Complex H\u2082O\u2082 Decomposition with Air<\/strong><\/h2>\n\n\n\n

Statement:<\/strong>
10 mL of a hydrogen peroxide solution is mixed with 350 mL of air containing 21% O\u2082 by volume. After complete catalytic decomposition of H\u2082O\u2082, the total volume of gases at the same temperature and pressure is 378 mL. Determine the volume strength of the original H\u2082O\u2082 solution.<\/p>\n\n\n\n

Solution:<\/strong>
Let the volume strength be ( x ) V.
From 10 mL H\u2082O\u2082, O\u2082 produced = ( 10x ) mL.<\/p>\n\n\n\n

Initially:
O\u2082 in air = ( 0.21 \\times 350 = 73.5 ) mL
N\u2082 in air = ( 350 – 73.5 = 276.5 ) mL (inert)<\/p>\n\n\n\n

After decomposition:
Additional O\u2082 from H\u2082O\u2082 = ( 10x ) mL
Total O\u2082 = ( 73.5 + 10x ) mL
N\u2082 unchanged = 276.5 mL
Total gas volume = ( (73.5 + 10x) + 276.5 = 350 + 10x ) mL<\/p>\n\n\n\n

Given:
$$
350 + 10x = 378 \\quad\\Rightarrow\\quad 10x = 28 \\quad\\Rightarrow\\quad x = 2.8
$$<\/p>\n\n\n\n

Answer:<\/strong> Volume strength = 2.8 V<\/strong><\/p>\n\n\n\n

\n\n\n\n

Problem 2: Mixing H\u2082O\u2082 Solutions with Different Strengths<\/strong><\/h2>\n\n\n\n

Statement:<\/strong>
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.<\/p>\n\n\n\n

Solution:<\/strong>
Let volumes mixed be ( V_1 ) mL (15 V) and ( V_2 ) mL (5 V).<\/p>\n\n\n\n

Moles of H\u2082O\u2082 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}
]<\/p>\n\n\n\n

Answer:<\/strong> Mixing ratio = 1:1<\/strong><\/p>\n\n\n\n

\n\n\n\n

Problem 3: Oleum Analysis with Percentage Labeling<\/strong><\/h2>\n\n\n\n

Statement:<\/strong>
An oleum sample labeled as “112% H\u2082SO\u2084” is given. This means 100 g of this oleum when treated with water gives 112 g of pure H\u2082SO\u2084. Calculate the percentage of free SO\u2083 in the oleum.<\/p>\n\n\n\n

Solution:<\/strong>
Let 100 g oleum contain ( x ) g free SO\u2083 and ( (100 – x) ) g H\u2082SO\u2084.<\/p>\n\n\n\n

Reaction: ( SO_3 + H_2O \\rightarrow H_2SO_4 )<\/p>\n\n\n\n

Moles of SO\u2083 = ( \\frac{x}{80} )
Mass of H\u2082SO\u2084 formed from SO\u2083 = ( \\frac{x}{80} \\times 98 = 1.225x ) g
Water consumed = ( \\frac{x}{80} \\times 18 = 0.225x ) g<\/p>\n\n\n\n

Total H\u2082SO\u2084 after hydration = ( (100 – x) + 1.225x = 100 + 0.225x ) g<\/p>\n\n\n\n

Given:
[
100 + 0.225x = 112 \\quad\\Rightarrow\\quad 0.225x = 12 \\quad\\Rightarrow\\quad x = \\frac{12}{0.225} = 53.33
]<\/p>\n\n\n\n

Answer:<\/strong> Free SO\u2083 = 53.33%<\/strong><\/p>\n\n\n\n

\n\n\n\n

Problem 4: Available Chlorine in Bleaching Powder Using Iodometric Titration<\/strong><\/h2>\n\n\n\n

Statement:<\/strong>
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 Na\u2082S\u2082O\u2083 solution. Calculate the percentage of available chlorine in the sample.<\/p>\n\n\n\n

Solution:<\/strong>
Reactions:
[
OCl^- + 2I^- + 2H^+ \\rightarrow I_2 + Cl^- + H_2O
]
[
I_2 + 2S_2O_3^{2-} \\rightarrow 2I^- + S_4O_6^{2-}
]<\/p>\n\n\n\n

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 \u2261 0.01 mol Cl\u2082 (available chlorine)<\/p>\n\n\n\n

Mass of Cl\u2082 = ( 0.01 \\times 71 = 0.71 ) g
[
\\% \\text{ available chlorine} = \\frac{0.71}{2.0} \\times 100 = 35.5\\%
]<\/p>\n\n\n\n

Answer:<\/strong> Available chlorine = 35.5%<\/strong><\/p>\n\n\n\n

\n\n\n\n

Problem 5: Combined Analysis of H\u2082O\u2082 and <\/strong><\/h2>\n\n\n\n

<\/p>\n","protected":false},"excerpt":{"rendered":"

Problem 1: Complex H\u2082O\u2082 Decomposition with Air Statement:10 mL of a hydrogen peroxide solution is mixed with 350 mL of air containing 21% O\u2082 by volume. After complete catalytic decomposition of H\u2082O\u2082, the total volume of gases at the same temperature and pressure is 378 mL. Determine the volume strength of the original H\u2082O\u2082 solution. […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":"","footnotes":""},"categories":[1],"tags":[],"class_list":["post-162834","post","type-post","status-publish","format-standard","hentry","category-uncategorized","cat-1-id"],"yoast_head":"\nJEE Advanced Numericals on VOLUME STRENGTH OF HYDROGEN PEROXIDE - Gyankatta<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/news.gyankatta.org\/?p=162834\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"JEE Advanced Numericals on VOLUME STRENGTH OF HYDROGEN PEROXIDE - Gyankatta\" \/>\n<meta property=\"og:description\" content=\"Problem 1: Complex H\u2082O\u2082 Decomposition with Air Statement:10 mL of a hydrogen peroxide solution is mixed with 350 mL of air containing 21% O\u2082 by volume. 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