Oxygen and cyclopropane at partial pressures orf 570 torr and 170 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?

Options

A. `(170xx42)/(570xx32)=0.39`
B. `(170)/(24)//(170/(42)+(570)/(32))~~0.19`
C. `(170)/(740)=0.23`
D. `(170)/(570)=0.30`

Explanation

To solve the problem of finding the ratio of the number of moles of cyclopropane to the number of moles of oxygen when they are mixed in a gas cylinder, we can use Dalton's Law of Partial Pressures. Here’s a step-by-step solution:

### Step 1: Understand Dalton's Law of Partial Pressures
Dalton's Law states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of each gas. The partial pressure of each gas is proportional to its mole fraction in the mixture.

### Step 2: Define the Given Values
We are given:
- Partial pressure of oxygen (P_O2) = 570 torr
- Partial pressure of cyclopropane (P_C3H6) = 170 torr

### Step 3: Write the Relationship Using Partial Pressures
According to Dalton's Law, we can express the mole fractions in terms of partial pressures:
\[ \frac{P_{O2}}{P_{C3H6}} = \frac{X_{O2}}{X_{C3H6}} \]

### Step 4: Substitute the Given Values
Substituting the values of the partial pressures into the equation:
\[ \frac{570 \, \text{torr}}{170 \, \text{torr}} = \frac{X_{O2}}{X_{C3H6}} \]

### Step 5: Calculate the Ratio of Partial Pressures
Calculating the left side:
\[ \frac{570}{170} = 3.3529 \approx 3.35 \]

### Step 6: Relate Mole Fractions to Moles
From the mole fraction definitions:
- \(X_{O2} = \frac{n_{O2}}{n_{total}}\)
- \(X_{C3H6} = \frac{n_{C3H6}}{n_{total}}\)

Thus, we can express the ratio of moles:
\[ \frac{X_{O2}}{X_{C3H6}} = \frac{n_{O2}}{n_{C3H6}} \]

### Step 7: Set Up the Ratio of Moles
From the previous steps, we can set up the ratio:
\[ \frac{n_{O2}}{n_{C3H6}} = \frac{P_{O2}}{P_{C3H6}} = \frac{570}{170} \]

### Step 8: Simplify the Ratio
Simplifying the ratio:
\[ \frac{n_{O2}}{n_{C3H6}} = \frac{570 \div 170}{170 \div 170} = \frac{3.35}{1} \]

### Step 9: Find the Ratio of Cyclopropane to Oxygen
To find the ratio of moles of cyclopropane to moles of oxygen, we take the reciprocal:
\[ \frac{n_{C3H6}}{n_{O2}} = \frac{1}{3.35} \approx 0.298 \]

### Final Answer
Thus, the ratio of the number of moles of cyclopropane to the number of moles of oxygen is approximately:
\[ \frac{n_{C3H6}}{n_{O2}} \approx 0.298 \text{ or } 1:3.35 \]