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A given volume of ozonised oxygen (containing 60% oxygen by volume) required 220 sec to effuse which an equal volume of oxygen took 200 sec only under the conditions. If density of O2 is 1.6 g/L then find density of O3:
- a)1.936 g/L
- b)2.16 g/L
- c)3.28 g/L
- d)2.24 g/L
Correct answer is option 'A'. Can you explain this answer?
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A given volume of ozonised oxygen (containing 60% oxygen by volume) re...
**Given Information:**
- The volume of ozonised oxygen required 220 sec to effuse.
- An equal volume of oxygen took 200 sec to effuse.
- The ozonised oxygen contains 60% oxygen by volume.
- The density of O2 is 1.6 g/L.
**To find:**
Density of O3.
**Solution:**
Effusion is the process of a gas escaping through a small hole into a vacuum. According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
**Step 1: Calculate the molar mass of O2 and O3:**
- Molar mass of O2 = 32 g/mol (16 g/mol per oxygen atom)
- Molar mass of O3 = 48 g/mol (16 g/mol per oxygen atom)
**Step 2: Calculate the ratio of effusion rates:**
- Let the effusion rate of O2 be R2 and the effusion rate of O3 be R3.
- According to Graham's law, R2/R3 = sqrt(M3/M2), where M2 and M3 are the molar masses of O2 and O3, respectively.
- Substituting the values, we get R2/R3 = sqrt(48/32) = sqrt(3/2).
**Step 3: Calculate the time ratio:**
- The time ratio of effusion is given as T2/T3 = R2/R3.
- Substituting the value of R2/R3, we get T2/T3 = sqrt(3/2).
**Step 4: Calculate the volume ratio:**
- The volume ratio is given as V2/V3 = T3/T2 (since equal volumes are being compared).
- Substituting the value of T2/T3, we get V2/V3 = 1/sqrt(3/2) = sqrt(2/3).
**Step 5: Calculate the molar ratio:**
- The molar ratio is given as n2/n3 = V2/V3, where n2 and n3 are the number of moles of O2 and O3, respectively.
- Substituting the value of V2/V3, we get n2/n3 = sqrt(2/3).
**Step 6: Calculate the volume ratio of oxygen in ozonised oxygen:**
- The volume ratio of oxygen in ozonised oxygen is given as V(O2)/V(O3) = 0.6 (since it contains 60% oxygen by volume).
- The volume ratio of O2/O3 can be calculated as V(O2)/V(O3) = 0.6 / (1 - 0.6) = 0.6 / 0.4 = 1.5.
**Step 7: Calculate the molar ratio of oxygen in ozonised oxygen:**
- The molar ratio of oxygen in ozonised oxygen is given as n(O2)/n(O3) = V(O2)/V(O3).
- Substituting the value of V(O2)/V(O3), we get n(O2)/n(O3) = 1.5.
**Step 8: Calculate the molar mass of ozonised oxygen:**
- Let the molar mass
- The volume of ozonised oxygen required 220 sec to effuse.
- An equal volume of oxygen took 200 sec to effuse.
- The ozonised oxygen contains 60% oxygen by volume.
- The density of O2 is 1.6 g/L.
**To find:**
Density of O3.
**Solution:**
Effusion is the process of a gas escaping through a small hole into a vacuum. According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
**Step 1: Calculate the molar mass of O2 and O3:**
- Molar mass of O2 = 32 g/mol (16 g/mol per oxygen atom)
- Molar mass of O3 = 48 g/mol (16 g/mol per oxygen atom)
**Step 2: Calculate the ratio of effusion rates:**
- Let the effusion rate of O2 be R2 and the effusion rate of O3 be R3.
- According to Graham's law, R2/R3 = sqrt(M3/M2), where M2 and M3 are the molar masses of O2 and O3, respectively.
- Substituting the values, we get R2/R3 = sqrt(48/32) = sqrt(3/2).
**Step 3: Calculate the time ratio:**
- The time ratio of effusion is given as T2/T3 = R2/R3.
- Substituting the value of R2/R3, we get T2/T3 = sqrt(3/2).
**Step 4: Calculate the volume ratio:**
- The volume ratio is given as V2/V3 = T3/T2 (since equal volumes are being compared).
- Substituting the value of T2/T3, we get V2/V3 = 1/sqrt(3/2) = sqrt(2/3).
**Step 5: Calculate the molar ratio:**
- The molar ratio is given as n2/n3 = V2/V3, where n2 and n3 are the number of moles of O2 and O3, respectively.
- Substituting the value of V2/V3, we get n2/n3 = sqrt(2/3).
**Step 6: Calculate the volume ratio of oxygen in ozonised oxygen:**
- The volume ratio of oxygen in ozonised oxygen is given as V(O2)/V(O3) = 0.6 (since it contains 60% oxygen by volume).
- The volume ratio of O2/O3 can be calculated as V(O2)/V(O3) = 0.6 / (1 - 0.6) = 0.6 / 0.4 = 1.5.
**Step 7: Calculate the molar ratio of oxygen in ozonised oxygen:**
- The molar ratio of oxygen in ozonised oxygen is given as n(O2)/n(O3) = V(O2)/V(O3).
- Substituting the value of V(O2)/V(O3), we get n(O2)/n(O3) = 1.5.
**Step 8: Calculate the molar mass of ozonised oxygen:**
- Let the molar mass
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Community Answer
A given volume of ozonised oxygen (containing 60% oxygen by volume) re...
Let VmL of gas effused.
V/220V/200=√dO2/dmix
⇒dmix=1.6×(1.1)^2=1.936g/L
Let density of ozone is d; In 100 volume ozonised oxygen, 60%O2 and 40% by volume O3 is present.
∴ Mass of mixture=mass of ozone + mass of oxygen
100×1.936=40×d+60×1.6
Density of O3(d)=2.44g/L.
V/220V/200=√dO2/dmix
⇒dmix=1.6×(1.1)^2=1.936g/L
Let density of ozone is d; In 100 volume ozonised oxygen, 60%O2 and 40% by volume O3 is present.
∴ Mass of mixture=mass of ozone + mass of oxygen
100×1.936=40×d+60×1.6
Density of O3(d)=2.44g/L.
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A given volume of ozonised oxygen (containing 60% oxygen by volume) required 220 sec to effuse which an equal volume of oxygen took 200 sec only under the conditions. If density of O2 is 1.6 g/L then find density of O3:a)1.936 g/Lb)2.16 g/Lc)3.28 g/Ld)2.24 g/LCorrect answer is option 'A'. Can you explain this answer?
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A given volume of ozonised oxygen (containing 60% oxygen by volume) required 220 sec to effuse which an equal volume of oxygen took 200 sec only under the conditions. If density of O2 is 1.6 g/L then find density of O3:a)1.936 g/Lb)2.16 g/Lc)3.28 g/Ld)2.24 g/LCorrect answer is option 'A'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A given volume of ozonised oxygen (containing 60% oxygen by volume) required 220 sec to effuse which an equal volume of oxygen took 200 sec only under the conditions. If density of O2 is 1.6 g/L then find density of O3:a)1.936 g/Lb)2.16 g/Lc)3.28 g/Ld)2.24 g/LCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A given volume of ozonised oxygen (containing 60% oxygen by volume) required 220 sec to effuse which an equal volume of oxygen took 200 sec only under the conditions. If density of O2 is 1.6 g/L then find density of O3:a)1.936 g/Lb)2.16 g/Lc)3.28 g/Ld)2.24 g/LCorrect answer is option 'A'. Can you explain this answer?.
A given volume of ozonised oxygen (containing 60% oxygen by volume) required 220 sec to effuse which an equal volume of oxygen took 200 sec only under the conditions. If density of O2 is 1.6 g/L then find density of O3:a)1.936 g/Lb)2.16 g/Lc)3.28 g/Ld)2.24 g/LCorrect answer is option 'A'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A given volume of ozonised oxygen (containing 60% oxygen by volume) required 220 sec to effuse which an equal volume of oxygen took 200 sec only under the conditions. If density of O2 is 1.6 g/L then find density of O3:a)1.936 g/Lb)2.16 g/Lc)3.28 g/Ld)2.24 g/LCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A given volume of ozonised oxygen (containing 60% oxygen by volume) required 220 sec to effuse which an equal volume of oxygen took 200 sec only under the conditions. If density of O2 is 1.6 g/L then find density of O3:a)1.936 g/Lb)2.16 g/Lc)3.28 g/Ld)2.24 g/LCorrect answer is option 'A'. Can you explain this answer?.
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