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The degree of dissociation of HI at a particular temperature is 0.8, calculate the volume of 2M Na2S2O3(soln) required to reduce completely the I2 present in the equilibrium mixture which was achieved by reacting 2 mole of each H2 and I2 in a vessel of volume 2L to produce HI.?
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Calculation of Volume of Na2S2O3 Solution Required to Reduce I2

Given:
- Degree of dissociation of HI (α) = 0.8
- Concentration of Na2S2O3 solution = 2M
- Volume of reaction vessel = 2L

To calculate the volume of Na2S2O3 solution required to completely reduce the I2 present in the equilibrium mixture, we need to first understand the reaction that takes place.

The reaction involved is as follows:
H2 + I2 ⇌ 2HI

From the balanced equation, we can see that 1 mole of I2 reacts with 1 mole of H2 to produce 2 moles of HI. Therefore, in the given reaction vessel with a volume of 2L, we initially have 2 moles each of H2 and I2.

Step 1: Calculate the moles of I2 remaining in the equilibrium mixture

Since the degree of dissociation of HI is given as 0.8, it means that 80% of the HI has dissociated into H2 and I2. Therefore, the remaining HI concentration is 20% or 0.2.

Initially, the concentration of HI is 2 moles / 2L = 1M. After dissociation, the concentration of HI becomes 0.2M.

Since 1 mole of HI reacts with 1 mole of I2, the moles of I2 remaining in the equilibrium mixture can be calculated as follows:
Moles of I2 = Moles of HI = 0.2M * 2L = 0.4 moles

Step 2: Calculate the volume of Na2S2O3 solution required to completely reduce the I2

In the reaction between I2 and Na2S2O3, each mole of I2 reacts with 2 moles of Na2S2O3. Therefore, to completely reduce the 0.4 moles of I2 remaining, we need:
Moles of Na2S2O3 = 2 * 0.4 = 0.8 moles

Now, we can calculate the volume of Na2S2O3 solution required:
Volume of Na2S2O3 solution = Moles of Na2S2O3 / Concentration of Na2S2O3
Volume of Na2S2O3 solution = 0.8 moles / 2M = 0.4L or 400mL

Therefore, the volume of 2M Na2S2O3 solution required to completely reduce the I2 present in the equilibrium mixture is 0.4L or 400mL.

Summary:
- Moles of I2 remaining in the equilibrium mixture = 0.4 moles
- Moles of Na2S2O3 required to reduce I2 = 0.8 moles
- Volume of Na2S2O3 solution required = 0.4L or 400mL
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The degree of dissociation of HI at a particular temperature is 0.8, calculate the volume of 2M Na2S2O3(soln) required to reduce completely the I2 present in the equilibrium mixture which was achieved by reacting 2 mole of each H2 and I2 in a vessel of volume 2L to produce HI.?
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The degree of dissociation of HI at a particular temperature is 0.8, calculate the volume of 2M Na2S2O3(soln) required to reduce completely the I2 present in the equilibrium mixture which was achieved by reacting 2 mole of each H2 and I2 in a vessel of volume 2L to produce HI.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The degree of dissociation of HI at a particular temperature is 0.8, calculate the volume of 2M Na2S2O3(soln) required to reduce completely the I2 present in the equilibrium mixture which was achieved by reacting 2 mole of each H2 and I2 in a vessel of volume 2L to produce HI.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The degree of dissociation of HI at a particular temperature is 0.8, calculate the volume of 2M Na2S2O3(soln) required to reduce completely the I2 present in the equilibrium mixture which was achieved by reacting 2 mole of each H2 and I2 in a vessel of volume 2L to produce HI.?.
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