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Q:A 5.0 mL of solution of H2O2 liberates 0.508 g of iodine from acidified KI solution.Calculate the strength of H2O2 solution in terms of volume strength at STP.?
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Q:A 5.0 mL of solution of H2O2 liberates 0.508 g of iodine from acidif...
Calculation of Volume Strength of H2O2 Solution
Given:
- Volume of H2O2 solution = 5.0 mL
- Mass of iodine liberated = 0.508 g
- Acidified KI solution used for the reaction
Step 1: Calculation of Number of moles of Iodine
The reaction between H2O2 and KI is as follows:
H2O2 + 2KI + 2H+ → I2 + 2K+ + 2H2O
From the reaction, it is clear that 1 mole of H2O2 liberates 1 mole of I2.
Therefore, the number of moles of I2 liberated can be calculated as follows:
Molar mass of I2 = 2 x atomic mass of I = 2 x 126.9 = 253.8 g/mol
Number of moles of I2 = Mass of I2 liberated / Molar mass of I2
= 0.508 / 253.8
= 0.002 mol
Step 2: Calculation of Volume Strength of H2O2 Solution
Volume strength is defined as the volume of H2O2 required to liberate 1 mL of oxygen at STP (Standard Temperature and Pressure).
1 mole of H2O2 liberates 1 mole of oxygen.
1 mole of any gas at STP occupies 22.4 L.
Therefore, the volume of oxygen liberated by 0.002 moles of H2O2 at STP can be calculated as follows:
Volume of oxygen liberated = 0.002 x 22.4
= 0.0448 L
= 44.8 mL
As 44.8 mL of oxygen is liberated by 0.002 moles of H2O2, the volume strength of H2O2 solution can be calculated as follows:
Volume strength of H2O2 solution = (44.8 / 5) x 1000
= 896 mL/L
= 896 volume strength
Therefore, the strength of H2O2 solution in terms of volume strength at STP is 896.
Given:
- Volume of H2O2 solution = 5.0 mL
- Mass of iodine liberated = 0.508 g
- Acidified KI solution used for the reaction
Step 1: Calculation of Number of moles of Iodine
The reaction between H2O2 and KI is as follows:
H2O2 + 2KI + 2H+ → I2 + 2K+ + 2H2O
From the reaction, it is clear that 1 mole of H2O2 liberates 1 mole of I2.
Therefore, the number of moles of I2 liberated can be calculated as follows:
Molar mass of I2 = 2 x atomic mass of I = 2 x 126.9 = 253.8 g/mol
Number of moles of I2 = Mass of I2 liberated / Molar mass of I2
= 0.508 / 253.8
= 0.002 mol
Step 2: Calculation of Volume Strength of H2O2 Solution
Volume strength is defined as the volume of H2O2 required to liberate 1 mL of oxygen at STP (Standard Temperature and Pressure).
1 mole of H2O2 liberates 1 mole of oxygen.
1 mole of any gas at STP occupies 22.4 L.
Therefore, the volume of oxygen liberated by 0.002 moles of H2O2 at STP can be calculated as follows:
Volume of oxygen liberated = 0.002 x 22.4
= 0.0448 L
= 44.8 mL
As 44.8 mL of oxygen is liberated by 0.002 moles of H2O2, the volume strength of H2O2 solution can be calculated as follows:
Volume strength of H2O2 solution = (44.8 / 5) x 1000
= 896 mL/L
= 896 volume strength
Therefore, the strength of H2O2 solution in terms of volume strength at STP is 896.
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Q:A 5.0 mL of solution of H2O2 liberates 0.508 g of iodine from acidif...
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Q:A 5.0 mL of solution of H2O2 liberates 0.508 g of iodine from acidified KI solution.Calculate the strength of H2O2 solution in terms of volume strength at STP.?
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Q:A 5.0 mL of solution of H2O2 liberates 0.508 g of iodine from acidified KI solution.Calculate the strength of H2O2 solution in terms of volume strength at STP.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Q:A 5.0 mL of solution of H2O2 liberates 0.508 g of iodine from acidified KI solution.Calculate the strength of H2O2 solution in terms of volume strength at STP.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Q:A 5.0 mL of solution of H2O2 liberates 0.508 g of iodine from acidified KI solution.Calculate the strength of H2O2 solution in terms of volume strength at STP.?.
Q:A 5.0 mL of solution of H2O2 liberates 0.508 g of iodine from acidified KI solution.Calculate the strength of H2O2 solution in terms of volume strength at STP.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Q:A 5.0 mL of solution of H2O2 liberates 0.508 g of iodine from acidified KI solution.Calculate the strength of H2O2 solution in terms of volume strength at STP.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Q:A 5.0 mL of solution of H2O2 liberates 0.508 g of iodine from acidified KI solution.Calculate the strength of H2O2 solution in terms of volume strength at STP.?.
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