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when 100ml of solution of H2O2 reacts completely with KI , then it oxidises KI into I2 . Now liberated iodine required 25ml of 1M Na2S2O3 in basic medium , where it oxidises into SO4 2negative , then what is the volume strength of H2O2
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when 100ml of solution of H2O2 reacts completely with KI , then it oxi...
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
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when 100ml of solution of H2O2 reacts completely with KI , then it oxi...
Calculation of Volume Strength of H2O2
Given Data:
- Initial volume of H2O2 solution = 100 ml
- Volume of Na2S2O3 required to react with liberated iodine = 25 ml
- Molarity of Na2S2O3 solution = 1 M
Step 1: Determining the moles of Iodine (I2) liberated
- From the reaction between H2O2 and KI, 1 mole of H2O2 liberates 1 mole of I2
- Volume of H2O2 solution = 100 ml = 0.1 L (as 1 L = 1000 ml)
- Number of moles of H2O2 = Molarity * Volume = 0.1 L * x M
- As per the reaction, moles of I2 = moles of H2O2 = 0.1x
Step 2: Determining the moles of Na2S2O3 required
- From the reaction between Na2S2O3 and I2, 1 mole of Na2S2O3 reacts with 1 mole of I2
- Moles of Na2S2O3 = Molarity * Volume = 1 * 0.025 L = 0.025 mol
Step 3: Calculation of Volume Strength of H2O2
- From Step 1, moles of I2 liberated = 0.1x
- From Step 2, moles of Na2S2O3 required = 0.025
- Since 1 mole of I2 reacts with 1 mole of Na2S2O3, moles of I2 = moles of Na2S2O3
- Therefore, 0.1x = 0.025
- Solving for x, we get x = 0.25 M
Result:
- The volume strength of the H2O2 solution is 0.25 M.
Given Data:
- Initial volume of H2O2 solution = 100 ml
- Volume of Na2S2O3 required to react with liberated iodine = 25 ml
- Molarity of Na2S2O3 solution = 1 M
Step 1: Determining the moles of Iodine (I2) liberated
- From the reaction between H2O2 and KI, 1 mole of H2O2 liberates 1 mole of I2
- Volume of H2O2 solution = 100 ml = 0.1 L (as 1 L = 1000 ml)
- Number of moles of H2O2 = Molarity * Volume = 0.1 L * x M
- As per the reaction, moles of I2 = moles of H2O2 = 0.1x
Step 2: Determining the moles of Na2S2O3 required
- From the reaction between Na2S2O3 and I2, 1 mole of Na2S2O3 reacts with 1 mole of I2
- Moles of Na2S2O3 = Molarity * Volume = 1 * 0.025 L = 0.025 mol
Step 3: Calculation of Volume Strength of H2O2
- From Step 1, moles of I2 liberated = 0.1x
- From Step 2, moles of Na2S2O3 required = 0.025
- Since 1 mole of I2 reacts with 1 mole of Na2S2O3, moles of I2 = moles of Na2S2O3
- Therefore, 0.1x = 0.025
- Solving for x, we get x = 0.25 M
Result:
- The volume strength of the H2O2 solution is 0.25 M.
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when 100ml of solution of H2O2 reacts completely with KI , then it oxidises KI into I2 . Now liberated iodine required 25ml of 1M Na2S2O3 in basic medium , where it oxidises into SO4 2negative , then what is the volume strength of H2O2
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when 100ml of solution of H2O2 reacts completely with KI , then it oxidises KI into I2 . Now liberated iodine required 25ml of 1M Na2S2O3 in basic medium , where it oxidises into SO4 2negative , then what is the volume strength of H2O2 for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about when 100ml of solution of H2O2 reacts completely with KI , then it oxidises KI into I2 . Now liberated iodine required 25ml of 1M Na2S2O3 in basic medium , where it oxidises into SO4 2negative , then what is the volume strength of H2O2 covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for when 100ml of solution of H2O2 reacts completely with KI , then it oxidises KI into I2 . Now liberated iodine required 25ml of 1M Na2S2O3 in basic medium , where it oxidises into SO4 2negative , then what is the volume strength of H2O2.
when 100ml of solution of H2O2 reacts completely with KI , then it oxidises KI into I2 . Now liberated iodine required 25ml of 1M Na2S2O3 in basic medium , where it oxidises into SO4 2negative , then what is the volume strength of H2O2 for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about when 100ml of solution of H2O2 reacts completely with KI , then it oxidises KI into I2 . Now liberated iodine required 25ml of 1M Na2S2O3 in basic medium , where it oxidises into SO4 2negative , then what is the volume strength of H2O2 covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for when 100ml of solution of H2O2 reacts completely with KI , then it oxidises KI into I2 . Now liberated iodine required 25ml of 1M Na2S2O3 in basic medium , where it oxidises into SO4 2negative , then what is the volume strength of H2O2.
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when 100ml of solution of H2O2 reacts completely with KI , then it oxidises KI into I2 . Now liberated iodine required 25ml of 1M Na2S2O3 in basic medium , where it oxidises into SO4 2negative , then what is the volume strength of H2O2, a detailed solution for when 100ml of solution of H2O2 reacts completely with KI , then it oxidises KI into I2 . Now liberated iodine required 25ml of 1M Na2S2O3 in basic medium , where it oxidises into SO4 2negative , then what is the volume strength of H2O2 has been provided alongside types of when 100ml of solution of H2O2 reacts completely with KI , then it oxidises KI into I2 . Now liberated iodine required 25ml of 1M Na2S2O3 in basic medium , where it oxidises into SO4 2negative , then what is the volume strength of H2O2 theory, EduRev gives you an
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