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Cyclopropane and oxygen at partial pressures 170 torr and 570 torr respectively are mixed in a gas cylinder. What is the ratio of the number of moles of cyclopropane to the number of moles of oxygen (n(C₃H₆/n(O₂))?

1. 170 × 42 / 570 × 32 ≈ 0.39
2. 170 / 570 × 42 / 32 ≈ 0.19
3. 170 / 740 ≈ 0.23
4. 170 / 570 ≈ 0.30

Correct answer: 170 / 570 × 42 / 32 ≈ 0.19

Solution

The ratio of moles is directly proportional to the ratio of partial pressures divided by their molar masses. Using the formula n(C₃H₆)/n(O₂) = (P(C₃H₆)/P(O₂)) × (M(O₂)/M(C₃H₆)), we get (170/570) × (32/42) ≈ 0.19.

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