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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.

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.

Grade:11

3 Answers

Kevin Nash
askIITians Faculty 332 Points
9 years ago
Sol. Meq of H2O2 = Meq of Na2S2O3 If N is normality of H2O2, then N × 25 = 0.3 × 20 ⇒ N = 0.24 ⇒ Volume strength = N × 5.6 = 1.334 V
Anshika
11 Points
5 years ago
M of H2O2 = M of Na2S2O3.If N = normality of H2O2; Then, N× 25 = 0.3 × 20N=0.24Now , volume strength=N × 5.6=0.24× 5.6 =》 1.344g/L
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem below.
 
Meq of H2O2 = Meq of Na2S2O3
If N is normality of H2O2, then
N × 25 = 0.3 × 20
⇒ N = 0.24
⇒ Volume strength = N × 5.6 = 1.334 V
 
Thanks and regards,
Kushagra
 

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