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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Redox Reactions

NTA Abhyas 2020 · Chemistry · Redox Reactions

To a 25 mL H 2 O 2 solution, excess of acidified solution of...

To a 25 mL H2O2 solution, excess of acidified solution of potassium iodide was added. The iodine liberated required 20 mL of 0.3 N sodium thiosulphate solution. Calculate the volume strength of H2O2 solution and report your answer by multiplying it with 1000.

Correct answer

1334

Step-by-step solution

Meq of H2O2 = Meq of I2 = Meq of Na2S2O3.

If N is normality of H2O2, then

N x 25 = 0.3 x 20, NH2O2=0.24N{\text{N}}_{{\text{H}}_{2} {\text{O}}_{2}} = 0 \text{.} 2 4 \text{N}

MH2O2=0.12 M{\text{M}}_{{\text{H}}_{2} {\text{O}}_{2}} = \text{0.12 M}

N = 0.24

Volume strength = 0.12×11.2\times \text{11.2}

=1.334 Vol

Final answer =1.334×1000=1334= 1.334 \times 1000 = 1334
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