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3L mixture of O2(g) and N2(g) is passed through ozonizer. The gas mixture obtained contains O2, O3 and N2 gases only. This obtained mixture is passed through alkaline pyrogallol and then through turpentine oil, there were 1100 mL and 600 mL volume contractions respectively. The volume of O2(g) and N2(g) in the original mixture respectively is -
  • a)
    1.5 L, 1.5 L
  • b)
    2.5 L, 0.5 L
  • c)
    1L, 2 L
  • d)
    2 L, 1 L
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
3L mixture of O2(g) and N2(g) is passed through ozonizer. The gas mixt...


t = 0   x = mL      0
t = tf   (x–3u)mL  (2u)mL= 600mL
∴ u = 300
⇒ x – 3 × 300 = 1100
∴ x = 2000 ∴ Eqn (1) ⇒ y = 1000 V
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Most Upvoted Answer
3L mixture of O2(g) and N2(g) is passed through ozonizer. The gas mixt...

Given Data:
- Volume contraction in alkaline pyrogallol: 1100 mL
- Volume contraction in turpentine oil: 600 mL

Calculations:
- Total volume contraction: 1100 mL + 600 mL = 1700 mL = 1.7 L

Let:
- Volume of O2 in the original mixture = x L
- Volume of N2 in the original mixture = 3x L (as the ratio of O2:N2 in air is 1:3)

Volume of gas components after passing through ozonizer:
- O2: x - 1.7 L
- O3: 0.3x L (as 30% of O2 gets converted to O3)
- N2: 3x - 1.7 L

Given volume contractions:
- Volume contraction in alkaline pyrogallol: 1100 mL = 1.1 L
- Volume contraction in turpentine oil: 600 mL = 0.6 L

Equation for volume contractions:
- O2 + O3 = 1.1 L
- O3 + N2 = 0.6 L

Solving the equations:
- From O2 + O3 = 1.1 L and O3 + N2 = 0.6 L, we get:
x - 1.7 + 0.3x = 1.1
0.3x + 3x - 1.7 = 0.6
Solving the above equations, we get x = 2 L and 3x = 6 L

Therefore, the volume of O2(g) and N2(g) in the original mixture respectively is 2 L and 6 L.
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3L mixture of O2(g) and N2(g) is passed through ozonizer. The gas mixture obtained contains O2, O3 and N2 gases only. This obtained mixture is passed through alkaline pyrogallol and then through turpentine oil, there were 1100 mL and 600 mL volume contractions respectively. The volume of O2(g) and N2(g) in the original mixture respectively is -a)1.5 L, 1.5 Lb)2.5 L, 0.5 Lc)1L, 2 Ld)2 L, 1 LCorrect answer is option 'D'. Can you explain this answer?
Question Description
3L mixture of O2(g) and N2(g) is passed through ozonizer. The gas mixture obtained contains O2, O3 and N2 gases only. This obtained mixture is passed through alkaline pyrogallol and then through turpentine oil, there were 1100 mL and 600 mL volume contractions respectively. The volume of O2(g) and N2(g) in the original mixture respectively is -a)1.5 L, 1.5 Lb)2.5 L, 0.5 Lc)1L, 2 Ld)2 L, 1 LCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 3L mixture of O2(g) and N2(g) is passed through ozonizer. The gas mixture obtained contains O2, O3 and N2 gases only. This obtained mixture is passed through alkaline pyrogallol and then through turpentine oil, there were 1100 mL and 600 mL volume contractions respectively. The volume of O2(g) and N2(g) in the original mixture respectively is -a)1.5 L, 1.5 Lb)2.5 L, 0.5 Lc)1L, 2 Ld)2 L, 1 LCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 3L mixture of O2(g) and N2(g) is passed through ozonizer. The gas mixture obtained contains O2, O3 and N2 gases only. This obtained mixture is passed through alkaline pyrogallol and then through turpentine oil, there were 1100 mL and 600 mL volume contractions respectively. The volume of O2(g) and N2(g) in the original mixture respectively is -a)1.5 L, 1.5 Lb)2.5 L, 0.5 Lc)1L, 2 Ld)2 L, 1 LCorrect answer is option 'D'. Can you explain this answer?.
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